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Trigonometric Equation Solver - dCode

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  • Trigonometric Equation Solver
  • Mathematics
  • Symbolic Computation
  • Trig Equation Solving

Answers to Questions (FAQ)

  • What is a trigonometric equation?

A trigonometric equation is a mathematical expression with an equality between two elements containing unknown variables and trigonometric functions (cos, sin, tan, etc.).

Example: $ \cos(x) = sin(\pi) $

  • How to solve a trigonometric equation?

dCode automatically solves the trigonometric equations (with the = equality symbol) and calculates the values of the unknowns.

All trig functions are supported: sin() , cos() , tan() , but also the trigonometry reciprocal functions acos() , arcsin() , and so on. as well as the hyperbolic functions cosh() , sinh() , etc.

Example: $ \sin(x) = 0 $ returns the solution $ x = 0 $ (radian)

Some equations will have an infinite number of solutions (modulo $ \pi $ or $ 2\pi $ or with constants $ c_i $)

All angles are in radian.

  • How to solve multiple trigonometric equations?

Multiple trig equations with the same variables can be combined with a logical AND operator: && or ⋀ .

Also, any new line return will be considered as a new equation.

  • How to solve a trig equation step by step?

The dCode solver does not display the calculation steps because they do not reflect the steps of a human thinking but to the steps of a machine thinking (bit-by-bit binary calculation operations) far from a hand-held resolution. dCode allows you to check a result.

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Trigonometric Equations Calculator

Get detailed solutions to your math problems with our trigonometric equations step-by-step calculator . practice your math skills and learn step by step with our math solver. check out all of our online calculators here .,  example,  solved problems,  difficult problems.

Solved example of trigonometric equations

The reciprocal sine function is cosecant: $\frac{1}{\csc(x)}=\sin(x)$

Move everything to the left hand side of the equation

Combining like terms $8\sin\left(x\right)$ and $-4\sin\left(x\right)$

Factor the polynomial $4\sin\left(x\right)-2$ by it's greatest common factor (GCF): $2$

Divide both sides of the equation by $2$

Simplifying the quotients

Divide $0$ by $2$

Intermediate steps

We need to isolate the dependent variable $x$, we can do that by simultaneously subtracting $-1$ from both sides of the equation

$x+0=x$, where $x$ is any expression

Multiply $-1$ times $-1$

Divide $1$ by $2$

The angles where the function $\sin\left(x\right)$ is $\frac{1}{2}$ are

The angles expressed in radians in the same order are equal to

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Related Concepts

Trigonometric functions

Module 9: Trigonometric Identities and Equations

Solving trigonometric equations, learning outcomes.

  • Solve equations involving a single trigonometric function.
  • Solve trigonometric equations using a calculator.
  • Solve trigonometric equations that involve factoring.
  • Solve trigonometric equations using fundamental identities.
  • Solve trigonometric equations with multiple angles.
  • Solve right triangle problems.

Solving Linear Trigonometric Equations in Sine and Cosine

Trigonometric equations are, as the name implies, equations that involve trigonometric functions. Similar in many ways to solving polynomial equations or rational equations, only specific values of the variable will be solutions, if there are solutions at all. Often we will solve a trigonometric equation over a specified interval. However, just as often, we will be asked to find all possible solutions, and as trigonometric functions are periodic, solutions are repeated within each period. In other words, trigonometric equations may have an infinite number of solutions. Additionally, like rational equations, the domain of the function must be considered before we assume that any solution is valid. The period of both the sine function and the cosine function is [latex]2\pi [/latex]. In other words, every [latex]2\pi [/latex] units, the y- values repeat. If we need to find all possible solutions, then we must add [latex]2\pi k[/latex], where [latex]k[/latex] is an integer, to the initial solution. Recall the rule that gives the format for stating all possible solutions for a function where the period is [latex]2\pi :[/latex]

There are similar rules for indicating all possible solutions for the other trigonometric functions. Solving trigonometric equations requires the same techniques as solving algebraic equations. We read the equation from left to right, horizontally, like a sentence. We look for known patterns, factor, find common denominators, and substitute certain expressions with a variable to make solving a more straightforward process. However, with trigonometric equations, we also have the advantage of using the identities we developed in the previous sections.

Example 1: Solving a Linear Trigonometric Equation Involving the Cosine Function

Find all possible exact solutions for the equation [latex]\cos \theta =\frac{1}{2}[/latex].

From the unit circle , we know that

[latex]\begin{gathered}\cos \theta =\frac{1}{2} \\ \theta =\frac{\pi }{3},\frac{5\pi }{3} \end{gathered}[/latex]

These are the solutions in the interval [latex]\left[0,2\pi \right][/latex]. All possible solutions are given by

[latex]\theta =\frac{\pi }{3}\pm 2k\pi \text{ and }\theta =\frac{5\pi }{3}\pm 2k\pi [/latex]

where [latex]k[/latex] is an integer.

Example 2: Solving a Linear Equation Involving the Sine Function

Find all possible exact solutions for the equation [latex]\sin t=\frac{1}{2}[/latex].

Solving for all possible values of t means that solutions include angles beyond the period of [latex]2\pi [/latex]. From the unit circle, we can see that the solutions are [latex]t=\frac{\pi }{6}[/latex] and [latex]t=\frac{5\pi }{6}[/latex]. But the problem is asking for all possible values that solve the equation. Therefore, the answer is

[latex]t=\frac{\pi }{6}\pm 2\pi k\text{ and }t=\frac{5\pi }{6}\pm 2\pi k[/latex]

How To: Given a trigonometric equation, solve using algebra.

  • Look for a pattern that suggests an algebraic property, such as the difference of squares or a factoring opportunity.
  • Substitute the trigonometric expression with a single variable, such as [latex]x[/latex] or [latex]u[/latex].
  • Solve the equation the same way an algebraic equation would be solved.
  • Substitute the trigonometric expression back in for the variable in the resulting expressions.
  • Solve for the angle.

Example 3: Solve the Trigonometric Equation in Linear Form

Solve the equation exactly: [latex]2\cos \theta -3=-5,0\le \theta <2\pi [/latex].

[latex]\begin{gathered}2\cos \theta -3=-5 \\ \cos \theta =-2 \\ \cos \theta =-1 \\ \theta =\pi \end{gathered}[/latex]

Solve exactly the following linear equation on the interval [latex]\left[0,2\pi \right):2\sin x+1=0[/latex].

[latex]x=\frac{7\pi }{6},\frac{11\pi }{6}[/latex]

Solve Trigonometric Equations Using a Calculator

Not all functions can be solved exactly using only the unit circle. When we must solve an equation involving an angle other than one of the special angles, we will need to use a calculator. Make sure it is set to the proper mode, either degrees or radians, depending on the criteria of the given problem.

Example 4: Using a Calculator to Solve a Trigonometric Equation Involving Sine

Use a calculator to solve the equation [latex]\sin \theta =0.8[/latex], where [latex]\theta [/latex] is in radians.

Make sure mode is set to radians. To find [latex]\theta [/latex], use the inverse sine function. On most calculators, you will need to push the 2 ND button and then the SIN button to bring up the [latex]{\sin }^{-1}[/latex] function. What is shown on the screen is [latex]{\sin}^{-1}[/latex](. The calculator is ready for the input within the parentheses. For this problem, we enter [latex]{\sin }^{-1}\left(0.8\right)[/latex], and press ENTER. Thus, to four decimals places,

[latex]{\sin }^{-1}\left(0.8\right)\approx 0.9273[/latex]

This is the solution in quadrant I. There is also a solution in quadrant II. To find this we subtract [latex]/pi - 0.9273 \approx 2.2143 [/latex]

The general solution is

[latex]\theta \approx 0.9273\pm 2\pi k \text{ and } \theta \approx 2.2143 \pm 2\pi k[/latex]

The angle measurement in degrees is

[latex]\begin{align} \theta &\approx {53.1}^{\circ } \\ \theta &\approx {180}^{\circ }-{53.1}^{\circ } \\ &\approx {126.9}^{\circ } \end{align}[/latex]

Analysis of the Solution

Note that a calculator will only return an angle in quadrants I or IV for the sine function, since that is the range of the inverse sine. The other angle is obtained by using [latex]\pi -\theta [/latex].

Example 5: Using a Calculator to Solve a Trigonometric Equation Involving Secant

Use a calculator to solve the equation [latex]\sec \theta =-4[/latex], giving your answer in radians.

We can begin with some algebra.

[latex]\begin{gathered}\sec \theta =-4\\ \frac{1}{\cos \theta }=-4\\ \cos \theta =-\frac{1}{4}\end{gathered}[/latex]

Check that the MODE is in radians. Now use the inverse cosine function.

[latex]\begin{gathered}{\cos }^{-1}\left(-\frac{1}{4}\right)\approx 1.8235 \\ \theta \approx 1.8235+2\pi k \end{gathered}[/latex]

Since [latex]\frac{\pi }{2}\approx 1.57[/latex] and [latex]\pi \approx 3.14[/latex], 1.8235 is between these two numbers, thus [latex]\theta \approx \text{1}\text{.8235}[/latex] is in quadrant II. Cosine is also negative in quadrant III. Note that a calculator will only return an angle in quadrants I or II for the cosine function, since that is the range of the inverse cosine.

Graph of angles theta =approx 1.8235, theta prime =approx pi - 1.8235 = approx 1.3181, and then theta prime = pi + 1.3181 = approx 4.4597

So, we also need to find the measure of the angle in quadrant III. In quadrant III, the reference angle is [latex]\theta \text{ }\text{ }\text{‘}\approx \pi -\text{1}\text{.8235}\approx \text{1}\text{.3181}\text{.}[/latex] The other solution in quadrant III is [latex]\theta \text{ }\text{ }\text{‘}\approx \pi +\text{1}\text{.3181}\approx \text{4}\text{.4597}\text{.}[/latex]

The solutions are [latex]\theta \approx 1.8235\pm 2\pi k[/latex] and [latex]\theta \approx 4.4597\pm 2\pi k[/latex].

Solve [latex]\cos \theta =-0.2[/latex].

[latex]\theta \approx 1.7722\pm 2\pi k[/latex] and [latex]\theta \approx 4.5110\pm 2\pi k[/latex]

Solving Equations Involving a Single Trigonometric Function

When we are given equations that involve only one of the six trigonometric functions, their solutions involve using algebraic techniques and the unit circle. We need to make several considerations when the equation involves trigonometric functions other than sine and cosine. Problems involving the reciprocals of the primary trigonometric functions need to be viewed from an algebraic perspective. In other words, we will write the reciprocal function, and solve for the angles using the function. Also, an equation involving the tangent function is slightly different from one containing a sine or cosine function. First, as we know, the period of tangent is [latex]\pi [/latex], not [latex]2\pi [/latex]. Further, the domain of tangent is all real numbers with the exception of odd integer multiples of [latex]\frac{\pi }{2}[/latex], unless, of course, a problem places its own restrictions on the domain.

Example 6: Solving a Problem Involving a Single Trigonometric Function

Solve the problem exactly: [latex]2{\sin }^{2}\theta -1=0,0\le \theta <2\pi [/latex].

As this problem is not easily factored, we will solve using the square root property. First, we use algebra to isolate [latex]\sin \theta [/latex]. Then we will find the angles.

[latex]\begin{gathered}2{\sin }^{2}\theta -1=0 \\ 2{\sin }^{2}\theta =1 \\ {\sin }^{2}\theta =\frac{1}{2} \\ \sqrt{{\sin }^{2}\theta }=\pm \sqrt{\frac{1}{2}} \\ \sin \theta =\pm \frac{1}{\sqrt{2}}=\pm \frac{\sqrt{2}}{2} \\ \theta =\frac{\pi }{4},\frac{3\pi }{4},\frac{5\pi }{4},\frac{7\pi }{4} \end{gathered}[/latex]

Example 7: Solving a Trigonometric Equation Involving Cosecant

Solve the following equation exactly: [latex]\csc \theta =-2,0\le \theta <4\pi [/latex].

We want all values of [latex]\theta [/latex] for which [latex]\csc \theta =-2[/latex] over the interval [latex]0\le \theta <4\pi [/latex].

[latex]\begin{gathered}\csc \theta =-2 \\ \frac{1}{\sin \theta }=-2 \\ \sin \theta =-\frac{1}{2} \\ \theta =\frac{7\pi }{6},\frac{11\pi }{6},\frac{19\pi }{6},\frac{23\pi }{6} \end{gathered}[/latex]

As [latex]\sin \theta =-\frac{1}{2}[/latex], notice that all four solutions are in the third and fourth quadrants.

Example 8: Solving an Equation Involving Tangent

Solve the equation exactly: [latex]\tan \left(\theta -\frac{\pi }{2}\right)=1,0\le \theta <2\pi [/latex].

Recall that the tangent function has a period of [latex]\pi [/latex]. On the interval [latex]\left[0,\pi \right)[/latex], and at the angle of [latex]\frac{\pi }{4}[/latex], the tangent has a value of 1. However, the angle we want is [latex]\left(\theta -\frac{\pi }{2}\right)[/latex]. Thus, if [latex]\tan \left(\frac{\pi }{4}\right)=1[/latex], then

[latex]\begin{gathered}\theta -\frac{\pi }{2}=\frac{\pi }{4}\\ \theta =\frac{3\pi }{4}\pm k\pi \end{gathered}[/latex]

Over the interval [latex]\left[0,2\pi \right)[/latex], we have two solutions:

[latex]\theta =\frac{3\pi }{4}\text{ and }\theta =\frac{3\pi }{4}+\pi =\frac{7\pi }{4}[/latex]

Find all solutions for [latex]\tan x=\sqrt{3}[/latex].

[latex]\frac{\pi }{3}\pm \pi k[/latex]

Example 9: Identify all Solutions to the Equation Involving Tangent

Identify all exact solutions to the equation [latex]2\left(\tan x+3\right)=5+\tan x,0\le x<2\pi [/latex].

We can solve this equation using only algebra. Isolate the expression [latex]\tan x[/latex] on the left side of the equals sign.

[latex]\begin{gathered} 2\left(\tan x\right)+2\left(3\right) =5+\tan x \\ 2\tan x+6 =5+\tan x \\ 2\tan x-\tan x =5 - 6 \\ \tan x =-1\end{gathered}[/latex]

There are two angles on the unit circle that have a tangent value of [latex]-1:\theta =\frac{3\pi }{4}[/latex] and [latex]\theta =\frac{7\pi }{4}[/latex].

Solving Trigonometric Equations in Quadratic Form

Solving a quadratic equation may be more complicated, but once again, we can use algebra as we would for any quadratic equation. Look at the pattern of the equation. Is there more than one trigonometric function in the equation, or is there only one? Which trigonometric function is squared? If there is only one function represented and one of the terms is squared, think about the standard form of a quadratic. Replace the trigonometric function with a variable such as [latex]x[/latex] or [latex]u[/latex]. If substitution makes the equation look like a quadratic equation, then we can use the same methods for solving quadratics to solve the trigonometric equations.

Example 10: Solving a Trigonometric Equation in Quadratic Form

Solve the equation exactly: [latex]{\cos }^{2}\theta +3\cos \theta -1=0,0\le \theta <2\pi[/latex].

We begin by using substitution and replacing cos [latex]\theta[/latex] with [latex]x[/latex]. It is not necessary to use substitution, but it may make the problem easier to solve visually. Let [latex]\cos \theta =x[/latex]. We have

[latex]{x}^{2}+3x - 1=0[/latex]

The equation cannot be factored, so we will use the quadratic formula [latex]x=\frac{-b\pm \sqrt{{b}^{2}-4ac}}{2a}[/latex].

[latex]\begin{align} x&=\frac{-3\pm \sqrt{{\left(-3\right)}^{2}-4\left(1\right)\left(-1\right)}}{2}&=\frac{-3\pm \sqrt{13}}{2} \end{align}[/latex]

Replace [latex]x[/latex] with [latex]\cos \theta[/latex], and solve. Thus,

[latex]\begin{gathered} \cos \theta =\frac{-3\pm \sqrt{13}}{2}\theta ={\cos }^{-1}\left(\frac{-3+\sqrt{13}}{2}\right)\end{gathered}[/latex]

Note that only the + sign is used. This is because we get an error when we solve [latex]\theta ={\cos }^{-1}\left(\frac{-3-\sqrt{13}}{2}\right)[/latex] on a calculator, since the domain of the inverse cosine function is [latex]\left[-1,1\right][/latex]. However, there is a second solution:

[latex]\begin{align}\theta &={\cos }^{-1}\left(\frac{-3+\sqrt{13}}{2}\right) \\ &\approx 1.26 \end{align}[/latex]

This terminal side of the angle lies in quadrant I. Since cosine is also positive in quadrant IV, the second solution is

[latex]\begin{align}\theta &=2\pi -{\cos }^{-1}\left(\frac{-3+\sqrt{13}}{2}\right) \\ &\approx 5.02 \end{align}[/latex]

Example 11: Solving a Trigonometric Equation in Quadratic Form by Factoring

Solve the equation exactly: [latex]2{\sin }^{2}\theta -5\sin \theta +3=0,0\le \theta \le 2\pi[/latex].

Using grouping, this quadratic can be factored. Either make the real substitution, [latex]\sin \theta =u[/latex], or imagine it, as we factor:

[latex]\begin{gathered}2{\sin }^{2}\theta -5\sin \theta +3=0 \\ \left(2\sin \theta -3\right)\left(\sin \theta -1\right)=0 \end{gathered}[/latex]

Now set each factor equal to zero.

[latex]\begin{gathered}2\sin \theta -3=0 \\ 2\sin \theta =3 \\ \sin \theta =\frac{3}{2} \\ \text{ } \\ \sin \theta -1=0 \\ \sin \theta =1 \end{gathered}[/latex]

Next solve for [latex]\theta :\sin \theta \ne \frac{3}{2}[/latex], as the range of the sine function is [latex]\left[-1,1\right][/latex]. However, [latex]\sin \theta =1[/latex], giving the solution [latex]\theta =\frac{\pi }{2}[/latex].

Make sure to check all solutions on the given domain as some factors have no solution.

Solve [latex]{\sin }^{2}\theta =2\cos \theta +2,0\le \theta \le 2\pi[/latex]. [Hint: Make a substitution to express the equation only in terms of cosine.]

[latex]\cos \theta =-1,\theta =\pi [/latex]

Example 12: Solving a Trigonometric Equation Using Algebra

Solve exactly:

[latex]2{\sin }^{2}\theta +\sin \theta =0;0\le \theta <2\pi[/latex]

This problem should appear familiar as it is similar to a quadratic. Let [latex]\sin \theta =x[/latex]. The equation becomes [latex]2{x}^{2}+x=0[/latex]. We begin by factoring:

[latex]\begin{gathered}2{x}^{2}+x=0\\ x\left(2x+1\right)=0\end{gathered}[/latex]

Set each factor equal to zero.

[latex]\begin{gathered}x=0 \\ 2x+1=0 \\ x=-\frac{1}{2} \end{gathered}[/latex]

Then, substitute back into the equation the original expression [latex]\sin \theta[/latex] for [latex]x[/latex]. Thus,

[latex]\begin{gathered}\sin \theta =0 \\ \theta =0,\pi \\ \text{ } \\ \sin \theta =-\frac{1}{2} \\ \theta =\frac{7\pi }{6},\frac{11\pi }{6} \end{gathered}[/latex]

The solutions within the domain [latex]0\le \theta <2\pi[/latex] are [latex]\theta =0,\pi ,\frac{7\pi }{6},\frac{11\pi }{6}[/latex].

If we prefer not to substitute, we can solve the equation by following the same pattern of factoring and setting each factor equal to zero.

[latex]\begin{gathered}2{\sin }^{2}\theta +\sin \theta =0 \\ \sin \theta \left(2\sin \theta +1\right)=0 \\ \sin \theta =0 \\ \theta =0,\pi \\ \text{ } \\ 2\sin \theta +1=0 \\ 2\sin \theta =-1 \\ \sin \theta =-\frac{1}{2} \\ \theta =\frac{7\pi }{6},\frac{11\pi }{6} \end{gathered}[/latex]

We can see the solutions on the graph in Figure 3. On the interval [latex]0\le \theta <2\pi[/latex], the graph crosses the x- axis four times, at the solutions noted. Notice that trigonometric equations that are in quadratic form can yield up to four solutions instead of the expected two that are found with quadratic equations. In this example, each solution (angle) corresponding to a positive sine value will yield two angles that would result in that value.

Graph of 2*(sin(theta))^2 + sin(theta) from 0 to 2pi. Zeros are at 0, pi, 7pi/6, and 11pi/6.

We can verify the solutions on the unit circle in Sum and Difference Identities as well.

Example 13: Solving a Trigonometric Equation Quadratic in Form

Solve the equation quadratic in form exactly: [latex]2{\sin }^{2}\theta -3\sin \theta +1=0,0\le \theta <2\pi[/latex].

We can factor using grouping. Solution values of [latex]\theta[/latex] can be found on the unit circle:

[latex]\begin{gathered}\left(2\sin \theta -1\right)\left(\sin \theta -1\right)=0 \\ 2\sin \theta -1=0 \\ \sin \theta =\frac{1}{2} \\ \theta =\frac{\pi }{6},\frac{5\pi }{6} \\ \text{ } \\ \sin \theta =1 \\ \theta =\frac{\pi }{2} \end{gathered}[/latex]

Solve the quadratic equation [latex]2{\cos }^{2}\theta +\cos \theta =0[/latex].

[latex]\frac{\pi }{2},\frac{2\pi }{3},\frac{4\pi }{3},\frac{3\pi }{2}[/latex]

Solving Trigonometric Equations Using Fundamental Identities

While algebra can be used to solve a number of trigonometric equations, we can also use the fundamental identities because they make solving equations simpler. Remember that the techniques we use for solving are not the same as those for verifying identities. The basic rules of algebra apply here, as opposed to rewriting one side of the identity to match the other side. In the next example, we use two identities to simplify the equation.

Example 14: Use Identities to Solve an Equation

Use identities to solve exactly the trigonometric equation over the interval [latex]0\le x<2\pi [/latex].

[latex]\cos x\cos \left(2x\right)+\sin x\sin \left(2x\right)=\frac{\sqrt{3}}{2}[/latex]

Notice that the left side of the equation is the difference formula for cosine.

[latex]\begin{align} \cos x\cos \left(2x\right)+\sin x\sin \left(2x\right)&=\frac{\sqrt{3}}{2} \\ \cos \left(x - 2x\right)&=\frac{\sqrt{3}}{2}&& \text{Difference formula for cosine} \\ \cos \left(-x\right)&=\frac{\sqrt{3}}{2}&& \text{Use the negative angle identity}. \\ \cos x&=\frac{\sqrt{3}}{2}\end{align}[/latex]

From the unit circle in Sum and Difference Identities, we see that [latex]\cos x=\frac{\sqrt{3}}{2}[/latex] when [latex]x=\frac{\pi }{6},\frac{11\pi }{6}[/latex].

Example 15: Solving the Equation Using a Double-Angle Formula

Solve the equation exactly using a double-angle formula: [latex]\cos \left(2\theta \right)=\cos \theta [/latex].

We have three choices of expressions to substitute for the double-angle of cosine. As it is simpler to solve for one trigonometric function at a time, we will choose the double-angle identity involving only cosine:

[latex]\begin{gathered}\cos \left(2\theta \right)=\cos \theta \\ 2{\cos }^{2}\theta -1=\cos \theta \\ 2{\cos }^{2}\theta -\cos \theta -1=0 \\ \left(2\cos \theta +1\right)\left(\cos \theta -1\right)=0 \\ 2\cos \theta +1=0 \\ \cos \theta =-\frac{1}{2} \\ \text{ } \\ \cos \theta -1=0 \\ \cos \theta =1 \end{gathered}[/latex]

So, if [latex]\cos \theta =-\frac{1}{2}[/latex], then [latex]\theta =\frac{2\pi }{3}\pm 2\pi k[/latex] and [latex]\theta =\frac{4\pi }{3}\pm 2\pi k[/latex]; if [latex]\cos \theta =1[/latex], then [latex]\theta =0\pm 2\pi k[/latex].

Example 16: Solving an Equation Using an Identity

Solve the equation exactly using an identity: [latex]3\cos \theta +3=2{\sin }^{2}\theta ,0\le \theta <2\pi [/latex].

If we rewrite the right side, we can write the equation in terms of cosine:

[latex]\begin{align} 3 cos\theta +3& ={2 sin}^{2}\theta \\ 3 cos\theta +3& =2\left(1-{\text{cos}}^{2}\theta \right) \\ 3 cos\theta +3& =2 - 2{\cos }^{2}\theta \\ 2{\cos }^{2}\theta +3 cos\theta +1& =0 \\ \left(2 cos\theta +1\right)\left(\cos \theta +1\right)& =0 \\ 2 cos\theta +1& =0 \\ \cos \theta & =-\frac{1}{2} \\ \theta & =\frac{2\pi }{3},\frac{4\pi }{3} \\ \text{ } \\ \cos \theta +1& =0 \\ \cos \theta & =-1 \\ \theta & =\pi\end{align}[/latex]

Our solutions are [latex]\theta =\frac{2\pi }{3},\frac{4\pi }{3},\pi [/latex].

Solving Trigonometric Equations with Multiple Angles

Sometimes it is not possible to solve a trigonometric equation with identities that have a multiple angle, such as [latex]\sin \left(2x\right)[/latex] or [latex]\cos \left(3x\right)[/latex]. When confronted with these equations, recall that [latex]y=\sin \left(2x\right)[/latex] is a horizontal compression by a factor of 2 of the function [latex]y=\sin x[/latex]. On an interval of [latex]2\pi [/latex], we can graph two periods of [latex]y=\sin \left(2x\right)[/latex], as opposed to one cycle of [latex]y=\sin x[/latex]. This compression of the graph leads us to believe there may be twice as many x -intercepts or solutions to [latex]\sin \left(2x\right)=0[/latex] compared to [latex]\sin x=0[/latex]. This information will help us solve the equation.

Example 17: Solving a Multiple Angle Trigonometric Equation

Solve exactly: [latex]\cos \left(2x\right)=\frac{1}{2}[/latex] on [latex]\left[0,2\pi \right)[/latex].

We can see that this equation is the standard equation with a multiple of an angle. If [latex]\cos \left(\alpha \right)=\frac{1}{2}[/latex], we know [latex]\alpha [/latex] is in quadrants I and IV. While [latex]\theta ={\cos }^{-1}\frac{1}{2}[/latex] will only yield solutions in quadrants I and II, we recognize that the solutions to the equation [latex]\cos \theta =\frac{1}{2}[/latex] will be in quadrants I and IV.

Therefore, the possible angles are [latex]\theta =\frac{\pi }{3}[/latex] and [latex]\theta =\frac{5\pi }{3}[/latex]. So, [latex]2x=\frac{\pi }{3}[/latex] or [latex]2x=\frac{5\pi }{3}[/latex], which means that [latex]x=\frac{\pi }{6}[/latex] or [latex]x=\frac{5\pi }{6}[/latex]. Does this make sense? Yes, because [latex]\cos \left(2\left(\frac{\pi }{6}\right)\right)=\cos \left(\frac{\pi }{3}\right)=\frac{1}{2}[/latex].

Are there any other possible answers? Let us return to our first step.

In quadrant I, [latex]2x=\frac{\pi }{3}[/latex], so [latex]x=\frac{\pi }{6}[/latex] as noted. Let us revolve around the circle again:

[latex]\begin{align} 2x&=\frac{\pi }{3}+2\pi \\ &=\frac{\pi }{3}+\frac{6\pi }{3} \\ &=\frac{7\pi }{3} \end{align}[/latex]

so [latex]x=\frac{7\pi }{6}[/latex].

One more rotation yields

[latex]\begin{align} 2x&=\frac{\pi }{3}+4\pi \\ &=\frac{\pi }{3}+\frac{12\pi }{3} \\ &=\frac{13\pi }{3} \end{align}[/latex]

[latex]x=\frac{13\pi }{6}>2\pi [/latex], so this value for [latex]x[/latex] is larger than [latex]2\pi [/latex], so it is not a solution on [latex]\left[0,2\pi \right)[/latex].

In quadrant IV, [latex]2x=\frac{5\pi }{3}[/latex], so [latex]x=\frac{5\pi }{6}[/latex] as noted. Let us revolve around the circle again:

[latex]\begin{align}2x&=\frac{5\pi }{3}+2\pi \\ &=\frac{5\pi }{3}+\frac{6\pi }{3} \\ &=\frac{11\pi }{3}\end{align}[/latex]

so [latex]x=\frac{11\pi }{6}[/latex].

[latex]\begin{align}2x&=\frac{5\pi }{3}+4\pi \\ &=\frac{5\pi }{3}+\frac{12\pi }{3} \\ &=\frac{17\pi }{3} \end{align}[/latex]

[latex]x=\frac{17\pi }{6}>2\pi [/latex], so this value for [latex]x[/latex] is larger than [latex]2\pi [/latex], so it is not a solution on [latex]\left[0,2\pi \right)[/latex].

Our solutions are [latex]x=\frac{\pi }{6},\frac{5\pi }{6},\frac{7\pi }{6},\text{and }\frac{11\pi }{6}[/latex]. Note that whenever we solve a problem in the form of [latex]\sin \left(nx\right)=c[/latex], we must go around the unit circle [latex]n[/latex] times.

Solving Right Triangle Problems

We can now use all of the methods we have learned to solve problems that involve applying the properties of right triangles and the Pythagorean Theorem . We begin with the familiar Pythagorean Theorem, [latex]{a}^{2}+{b}^{2}={c}^{2}[/latex], and model an equation to fit a situation.

Example 18: Using the Pythagorean Theorem to Model an Equation

Use the Pythagorean Theorem, and the properties of right triangles to model an equation that fits the problem.

One of the cables that anchors the center of the London Eye Ferris wheel to the ground must be replaced. The center of the Ferris wheel is 69.5 meters above the ground, and the second anchor on the ground is 23 meters from the base of the Ferris wheel. Approximately how long is the cable, and what is the angle of elevation (from ground up to the center of the Ferris wheel)?

Basic diagram of a ferris wheel (circle) and its support cables (form a right triangle). One cable runs from the center of the circle to the ground (outside the circle), is perpendicular to the ground, and has length 69.5. Another cable of unknown length (the hypotenuse) runs from the center of the circle to the ground 23 feet away from the other cable at an angle of theta degrees with the ground. So, in closing, there is a right triangle with base 23, height 69.5, hypotenuse unknown, and angle between base and hypotenuse of theta degrees.

Using the information given, we can draw a right triangle. We can find the length of the cable with the Pythagorean Theorem.

[latex]\begin{gathered}{a}^{2}+{b}^{2}={c}^{2} \\ {\left(23\right)}^{2}+{\left(69.5\right)}^{2}\approx 5359 \\ \sqrt{5359}\approx 73.2\text{ m}\end{gathered}[/latex]

The angle of elevation is [latex]\theta [/latex], formed by the second anchor on the ground and the cable reaching to the center of the wheel. We can use the tangent function to find its measure. Round to two decimal places.

[latex]\begin{align}\tan \theta &=\frac{69.5}{23} \\ {\tan }^{-1}\left(\frac{69.5}{23}\right)&\approx 1.2522 \\ &\approx {71.69}^{\circ } \end{align}[/latex]

The angle of elevation is approximately [latex]{71.7}^{\circ }[/latex], and the length of the cable is 73.2 meters.

Example 19: Using the Pythagorean Theorem to Model an Abstract Problem

OSHA safety regulations require that the base of a ladder be placed 1 foot from the wall for every 4 feet of ladder length. Find the angle that a ladder of any length forms with the ground and the height at which the ladder touches the wall.

For any length of ladder, the base needs to be a distance from the wall equal to one fourth of the ladder’s length. Equivalently, if the base of the ladder is “ a” feet from the wall, the length of the ladder will be 4 a feet.

Diagram of a right triangle with base length a, height length b, hypotenuse length 4a. Opposite the height is an angle of theta degrees, and opposite the hypotenuse is an angle of 90 degrees.

The side adjacent to [latex]\theta [/latex] is a and the hypotenuse is [latex]4a[/latex]. Thus,

[latex]\begin{gathered}\cos \theta =\frac{a}{4a}=\frac{1}{4} \\ {\cos }^{-1}\left(\frac{1}{4}\right)\approx {75.5}^{\circ } \end{gathered}[/latex]

The elevation of the ladder forms an angle of [latex]{75.5}^{\circ }[/latex] with the ground. The height at which the ladder touches the wall can be found using the Pythagorean Theorem:

[latex]\begin{gathered}{a}^{2}+{b}^{2}={\left(4a\right)}^{2} \\ {b}^{2}={\left(4a\right)}^{2}-{a}^{2} \\ {b}^{2}=16{a}^{2}-{a}^{2} \\ {b}^{2}=15{a}^{2} \\ b=\sqrt{15}a \end{gathered}[/latex]

Thus, the ladder touches the wall at [latex]\sqrt{15}a[/latex] feet from the ground.

Key Concepts

  • When solving linear trigonometric equations, we can use algebraic techniques just as we do solving algebraic equations. Look for patterns, like the difference of squares, quadratic form, or an expression that lends itself well to substitution.
  • Equations involving a single trigonometric function can be solved or verified using the unit circle.
  • We can also solve trigonometric equations using a graphing calculator.
  • Many equations appear quadratic in form. We can use substitution to make the equation appear simpler, and then use the same techniques we use solving an algebraic quadratic: factoring, the quadratic formula, etc.
  • We can also use the identities to solve trigonometric equation.
  • We can use substitution to solve a multiple-angle trigonometric equation, which is a compression of a standard trigonometric function. We will need to take the compression into account and verify that we have found all solutions on the given interval.
  • Real-world scenarios can be modeled and solved using the Pythagorean Theorem and trigonometric functions.
  • Precalculus. Authored by : OpenStax College. Provided by : OpenStax. Located at : http://cnx.org/contents/[email protected]:1/Preface . License : CC BY: Attribution

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3.6: Solving Trigonometric Equations

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Learning Objectives

  • Solve linear trigonometric equations in sine and cosine.
  • Solve equations involving a single trigonometric function.
  • Solve trigonometric equations using a calculator.
  • Solve trigonometric equations that are quadratic in form.
  • Solve trigonometric equations using fundamental identities.
  • Solve trigonometric equations with multiple angles.
  • Solve right triangle problems.

Thales of Miletus (circa 625–547 BC) is known as the founder of geometry. The legend is that he calculated the height of the Great Pyramid of Giza in Egypt using the theory of similar triangles , which he developed by measuring the shadow of his staff. Based on proportions, this theory has applications in a number of areas, including fractal geometry, engineering, and architecture. Often, the angle of elevation and the angle of depression are found using similar triangles.

Photo of the Egyptian pyramids near a modern city.

In earlier sections of this chapter, we looked at trigonometric identities. Identities are true for all values in the domain of the variable. In this section, we begin our study of trigonometric equations to study real-world scenarios such as the finding the dimensions of the pyramids.

Solving Linear Trigonometric Equations in Sine and Cosine

Trigonometric equations are, as the name implies, equations that involve trigonometric functions. Similar in many ways to solving polynomial equations or rational equations, only specific values of the variable will be solutions, if there are solutions at all. Often we will solve a trigonometric equation over a specified interval. However, just as often, we will be asked to find all possible solutions, and as trigonometric functions are periodic, solutions are repeated within each period. In other words, trigonometric equations may have an infinite number of solutions. Additionally, like rational equations, the domain of the function must be considered before we assume that any solution is valid. The period of both the sine function and the cosine function is \(2\pi\). In other words, every \(2\pi\) units, the y- values repeat. If we need to find all possible solutions, then we must add \(2\pi k\),where \(k\) is an integer, to the initial solution. Recall the rule that gives the format for stating all possible solutions for a function where the period is \(2\pi\):

\[\sin \theta=\sin(\theta \pm 2k\pi)\]

There are similar rules for indicating all possible solutions for the other trigonometric functions. Solving trigonometric equations requires the same techniques as solving algebraic equations. We read the equation from left to right, horizontally, like a sentence. We look for known patterns, factor, find common denominators, and substitute certain expressions with a variable to make solving a more straightforward process. However, with trigonometric equations, we also have the advantage of using the identities we developed in the previous sections.

Example \(\PageIndex{1A}\): Solving a Linear Trigonometric Equation Involving the Cosine Function

Find all possible exact solutions for the equation \(\cos \theta=\dfrac{1}{2}\).

From the unit circle, we know that

\[ \begin{align*} \cos \theta &=\dfrac{1}{2} \\[4pt] \theta &=\dfrac{\pi}{3},\space \dfrac{5\pi}{3} \end{align*}\]

These are the solutions in the interval \([ 0,2\pi ]\). All possible solutions are given by

\[\theta=\dfrac{\pi}{3} \pm 2k\pi \quad \text{and} \quad \theta=\dfrac{5\pi}{3} \pm 2k\pi \nonumber\]

where \(k\) is an integer.

Example \(\PageIndex{1B}\): Solving a Linear Equation Involving the Sine Function

Find all possible exact solutions for the equation \(\sin t=\dfrac{1}{2}\).

Solving for all possible values of \(t\) means that solutions include angles beyond the period of \(2\pi\). From the section on Sum and Difference Identities , we can see that the solutions are \(t=\dfrac{\pi}{6}\) and \(t=\dfrac{5\pi}{6}\). But the problem is asking for all possible values that solve the equation. Therefore, the answer is

\[t=\dfrac{\pi}{6}\pm 2\pi k \quad \text{and} \quad t=\dfrac{5\pi}{6}\pm 2\pi k \nonumber\]

Howto: Given a trigonometric equation, solve using algebra

  • Look for a pattern that suggests an algebraic property, such as the difference of squares or a factoring opportunity.
  • Substitute the trigonometric expression with a single variable, such as \(x\) or \(u\).
  • Solve the equation the same way an algebraic equation would be solved.
  • Substitute the trigonometric expression back in for the variable in the resulting expressions.
  • Solve for the angle.

Example \(\PageIndex{2}\): Solve the Linear Trigonometric Equation

Solve the equation exactly: \(2 \cos \theta−3=−5\), \(0≤\theta<2\pi\).

Use algebraic techniques to solve the equation.

\[\begin{align*} 2 \cos \theta-3&= -5\\ 2 \cos \theta&= -2\\ \cos \theta&= -1\\ \theta&= \pi \end{align*}\]

Exercise \(\PageIndex{2}\)

Solve exactly the following linear equation on the interval \([0,2\pi)\): \(2 \sin x+1=0\).

\(x=\dfrac{7\pi}{6},\space \dfrac{11\pi}{6}\)

Solving Equations Involving a Single Trigonometric Function

When we are given equations that involve only one of the six trigonometric functions, their solutions involve using algebraic techniques and the unit circle (see [link] ). We need to make several considerations when the equation involves trigonometric functions other than sine and cosine. Problems involving the reciprocals of the primary trigonometric functions need to be viewed from an algebraic perspective. In other words, we will write the reciprocal function, and solve for the angles using the function. Also, an equation involving the tangent function is slightly different from one containing a sine or cosine function. First, as we know, the period of tangent is \(\pi\),not \(2\pi\). Further, the domain of tangent is all real numbers with the exception of odd integer multiples of \(\dfrac{\pi}{2}\),unless, of course, a problem places its own restrictions on the domain.

Example \(\PageIndex{3A}\): Solving a Problem Involving a Single Trigonometric Function

Solve the problem exactly: \(2 {\sin}^2 \theta−1=0\), \(0≤\theta<2\pi\).

As this problem is not easily factored, we will solve using the square root property. First, we use algebra to isolate \(\sin \theta\). Then we will find the angles.

\[\begin{align*} 2 {\sin}^2 \theta-1&= 0\\ 2 {\sin}^2 \theta&= 1\\ {\sin}^2 \theta&= \dfrac{1}{2}\\ \sqrt{ {\sin}^2 \theta }&= \pm \sqrt{ \dfrac{1}{2} }\\ \sin \theta&= \pm \dfrac{1}{\sqrt{2}}\\ &= \pm \dfrac{\sqrt{2}}{2}\\ \theta&= \dfrac{\pi}{4}, \space \dfrac{3\pi}{4},\space \dfrac{5\pi}{4}, \space \dfrac{7\pi}{4} \end{align*}\]

Example \(\PageIndex{3B}\): Solving a Trigonometric Equation Involving Cosecant

Solve the following equation exactly: \(\csc \theta=−2\), \(0≤\theta<4\pi\).

We want all values of \(\theta\) for which \(\csc \theta=−2\) over the interval \(0≤\theta<4\pi\).

\[\begin{align*} \csc \theta&= -2\\ \dfrac{1}{\sin \theta}&= -2\\ \sin \theta&= -\dfrac{1}{2}\\ \theta&= \dfrac{7\pi}{6},\space \dfrac{11\pi}{6},\space \dfrac{19\pi}{6}, \space \dfrac{23\pi}{6} \end{align*}\]

As \(\sin \theta=−\dfrac{1}{2}\), notice that all four solutions are in the third and fourth quadrants.

Example \(\PageIndex{3C}\): Solving an Equation Involving Tangent

Solve the equation exactly: \(\tan\left(\theta−\dfrac{\pi}{2}\right)=1\), \(0≤\theta<2\pi\).

Recall that the tangent function has a period of \(\pi\). On the interval \([ 0,\pi )\),and at the angle of \(\dfrac{\pi}{4}\),the tangent has a value of \(1\). However, the angle we want is \(\left(\theta−\dfrac{\pi}{2}\right)\). Thus, if \(\tan\left(\dfrac{\pi}{4}\right)=1\),then

\[\begin{align*} \theta-\dfrac{\pi}{2}&= \dfrac{\pi}{4}\\ \theta&= \dfrac{3\pi}{4} \pm k\pi \end{align*}\]

Over the interval \([ 0,2\pi )\),we have two solutions:

\(\theta=\dfrac{3\pi}{4}\) and \(\theta=\dfrac{3\pi}{4}+\pi=\dfrac{7\pi}{4}\)

Exercise \(\PageIndex{3}\)

Find all solutions for \(\tan x=\sqrt{3}\).

\(\dfrac{\pi}{3}\pm \pi k\)

Example \(\PageIndex{4}\): Identify all Solutions to the Equation Involving Tangent

Identify all exact solutions to the equation \(2(\tan x+3)=5+\tan x\), \(0≤x<2\pi\).

We can solve this equation using only algebra. Isolate the expression \(\tan x\) on the left side of the equals sign.

\[\begin{align*} 2(\tan x)+2(3)&= 5+\tan x\\ 2\tan x+6&= 5+\tan x\\ 2\tan x-\tan x&= 5-6\\ \tan x&= -1 \end{align*}\]

There are two angles on the unit circle that have a tangent value of \(−1\): \(\theta=\dfrac{3\pi}{4}\) and \(\theta=\dfrac{7\pi}{4}\).

Solve Trigonometric Equations Using a Calculator

Not all functions can be solved exactly using only the unit circle. When we must solve an equation involving an angle other than one of the special angles, we will need to use a calculator. Make sure it is set to the proper mode, either degrees or radians, depending on the criteria of the given problem.

Example \(\PageIndex{5A}\): Using a Calculator to Solve a Trigonometric Equation Involving Sine

Use a calculator to solve the equation \(\sin \theta=0.8\),where \(\theta\) is in radians.

Make sure mode is set to radians. To find \(\theta\), use the inverse sine function. On most calculators, you will need to push the 2 ND button and then the SIN button to bring up the \({\sin}^{−1}\) function. What is shown on the screen is \({\sin}^{−1}\).The calculator is ready for the input within the parentheses. For this problem, we enter \({\sin}^{−1}(0.8)\), and press ENTER. Thus, to four decimals places,

\({\sin}^{−1}(0.8)≈0.9273\)

The solution is

\(\theta≈0.9273\pm 2\pi k\)

The angle measurement in degrees is

\[\begin{align*} \theta&\approx 53.1^{\circ}\\ \theta&\approx 180^{\circ}-53.1^{\circ}\\ &\approx 126.9^{\circ} \end{align*}\]

Note that a calculator will only return an angle in quadrants I or IV for the sine function, since that is the range of the inverse sine. The other angle is obtained by using \(\pi−\theta\).

Example \(\PageIndex{5B}\): Using a Calculator to Solve a Trigonometric Equation Involving Secant

Use a calculator to solve the equation \( \sec θ=−4, \) giving your answer in radians.

We can begin with some algebra.

\[\begin{align*} \sec \theta&= -4\\ \dfrac{1}{\cos \theta}&= -4\\ \cos \theta&= -\dfrac{1}{4} \end{align*}\]

Check that the MODE is in radians. Now use the inverse cosine function

\[\begin{align*}{\cos}^{-1}\left(-\dfrac{1}{4}\right)&\approx 1.8235\\ \theta&\approx 1.8235+2\pi k \end{align*}\]

Since \(\dfrac{\pi}{2}≈1.57\) and \(\pi≈3.14\),\(1.8235\) is between these two numbers, thus \(\theta≈1.8235\) is in quadrant II. Cosine is also negative in quadrant III. Note that a calculator will only return an angle in quadrants I or II for the cosine function, since that is the range of the inverse cosine. See Figure \(\PageIndex{2}\).

Graph of angles theta =approx 1.8235, theta prime =approx pi - 1.8235 = approx 1.3181, and then theta prime = pi + 1.3181 = approx 4.4597

So, we also need to find the measure of the angle in quadrant III. In quadrant III, the reference angle is \(\theta '≈\pi−1.8235≈1.3181\). The other solution in quadrant III is \(\theta '≈\pi+1.3181≈4.4597\).

The solutions are \(\theta≈1.8235\pm 2\pi k\) and \(\theta≈4.4597\pm 2\pi k\).

Exercise \(\PageIndex{5}\)

Solve \(\cos \theta=−0.2\).

\(\theta≈1.7722\pm 2\pi k\) and \(\theta≈4.5110\pm 2\pi k\)

Solving Trigonometric Equations in Quadratic Form

Solving a quadratic equation may be more complicated, but once again, we can use algebra as we would for any quadratic equation. Look at the pattern of the equation. Is there more than one trigonometric function in the equation, or is there only one? Which trigonometric function is squared? If there is only one function represented and one of the terms is squared, think about the standard form of a quadratic. Replace the trigonometric function with a variable such as \(x\) or \(u\). If substitution makes the equation look like a quadratic equation, then we can use the same methods for solving quadratics to solve the trigonometric equations.

Example \(\PageIndex{6A}\): Solving a Trigonometric Equation in Quadratic Form

Solve the equation exactly: \({\cos}^2 \theta+3 \cos \theta−1=0\), \(0≤\theta<2\pi\).

We begin by using substitution and replacing \(\cos \theta\) with \(x\). It is not necessary to use substitution, but it may make the problem easier to solve visually. Let \(\cos \theta=x\). We have

\(x^2+3x−1=0\)

The equation cannot be factored, so we will use the quadratic formula: \(x=\dfrac{−b\pm \sqrt{b^2−4ac}}{2a}\).

\[\begin{align*} x&= \dfrac{ -3\pm \sqrt{ {(-3)}^2-4 (1) (-1) } }{2}\\ &= \dfrac{-3\pm \sqrt{13}}{2}\end{align*}\]

Replace \(x\) with \(\cos \theta \) and solve.

\[\begin{align*} \cos \theta&= \dfrac{-3\pm \sqrt{13}}{2}\\ \theta&= {\cos}^{-1}\left(\dfrac{-3+\sqrt{13}}{2}\right) \end{align*}\]

Note that only the + sign is used. This is because we get an error when we solve \(\theta={\cos}^{−1}\left(\dfrac{−3−\sqrt{13}}{2}\right)\) on a calculator, since the domain of the inverse cosine function is \([ −1,1 ]\). However, there is a second solution:

\[\begin{align*} \theta&= {\cos}^{-1}\left(\dfrac{-3+\sqrt{13}}{2}\right)\\ &\approx 1.26 \end{align*}\]

This terminal side of the angle lies in quadrant I. Since cosine is also positive in quadrant IV, the second solution is

\[\begin{align*} \theta&= 2\pi-{\cos}^{-1}\left(\dfrac{-3+\sqrt{13}}{2}\right)\\ &\approx 5.02 \end{align*}\]

Example \(\PageIndex{6B}\): Solving a Trigonometric Equation in Quadratic Form by Factoring

Solve the equation exactly: \(2 {\sin}^2 \theta−5 \sin \theta+3=0\), \(0≤\theta≤2\pi\).

Using grouping, this quadratic can be factored. Either make the real substitution, \(\sin \theta=u\),or imagine it, as we factor:

\[\begin{align*} 2 {\sin}^2 \theta-5 \sin \theta+3&= 0\\ (2 \sin \theta-3)(\sin \theta-1)&= 0 \qquad \text {Now set each factor equal to zero.}\\ 2 \sin \theta-3&= 0\\ 2 \sin \theta&= 3\\ \sin \theta&= \dfrac{3}{2}\\ \sin \theta-1&= 0\\ \sin \theta&= 1 \end{align*}\]

Next solve for \(\theta\): \(\sin \theta≠\dfrac{3}{2}\), as the range of the sine function is \([ −1,1 ]\). However, \(\sin \theta=1\), giving the solution \(\theta=\dfrac{\pi}{2}\).

Make sure to check all solutions on the given domain as some factors have no solution.

Exercise \(\PageIndex{6}\)

Solve \({\sin}^2 \theta=2 \cos \theta+2\), \(0≤\theta≤2\pi\). [Hint: Make a substitution to express the equation only in terms of cosine.]

\(\cos \theta=−1\), \(\theta=\pi\)

Example \(\PageIndex{7A}\): Solving a Trigonometric Equation Using Algebra

Solve exactly: \(2 {\sin}^2 \theta+\sin \theta=0;\space 0≤\theta<2\pi\)

This problem should appear familiar as it is similar to a quadratic. Let \(\sin \theta=x\). The equation becomes \(2x^2+x=0\). We begin by factoring:

\[\begin{align*} 2x^2+x&= 0\\ x(2x+1)&= 0\qquad \text {Set each factor equal to zero.}\\ x&= 0\\ 2x+1&= 0\\ x&= -\dfrac{1}{2} \end{align*}\] Then, substitute back into the equation the original expression \(\sin \theta \) for \(x\). Thus, \[\begin{align*} \sin \theta&= 0\\ \theta&= 0,\pi\\ \sin \theta&= -\dfrac{1}{2}\\ \theta&= \dfrac{7\pi}{6},\dfrac{11\pi}{6} \end{align*}\]

The solutions within the domain \(0≤\theta<2\pi\) are \(\theta=0,\pi,\dfrac{7\pi}{6},\dfrac{11\pi}{6}\).

If we prefer not to substitute, we can solve the equation by following the same pattern of factoring and setting each factor equal to zero.

\[\begin{align*} {\sin}^2 \theta+\sin \theta&= 0\\ \sin \theta(2\sin \theta+1)&= 0\\ \sin \theta&= 0\\ \theta&= 0,\pi\\ 2 \sin \theta+1&= 0\\ 2\sin \theta&= -1\\ \sin \theta&= -\dfrac{1}{2}\\ \theta&= \dfrac{7\pi}{6},\dfrac{11\pi}{6} \end{align*}\]

We can see the solutions on the graph in Figure \(\PageIndex{3}\). On the interval \(0≤\theta<2\pi\),the graph crosses the \(x\) - axis four times, at the solutions noted. Notice that trigonometric equations that are in quadratic form can yield up to four solutions instead of the expected two that are found with quadratic equations. In this example, each solution (angle) corresponding to a positive sine value will yield two angles that would result in that value.

Graph of 2*(sin(theta))^2 + sin(theta) from 0 to 2pi. Zeros are at 0, pi, 7pi/6, and 11pi/6.

We can verify the solutions on the unit circle in via the result in the section on Sum and Difference Identities as well.

Example \(\PageIndex{7B}\): Solving a Trigonometric Equation Quadratic in Form

Solve the equation quadratic in form exactly: \(2 {\sin}^2 \theta−3 \sin \theta+1=0\), \(0≤\theta<2\pi\).

We can factor using grouping. Solution values of \(\theta\) can be found on the unit circle.

\[\begin{align*} (2 \sin \theta-1)(\sin \theta-1)&= 0\\ 2 \sin \theta-1&= 0\\ \sin \theta&= \dfrac{1}{2}\\ \theta&= \dfrac{\pi}{6}, \dfrac{5\pi}{6}\\ \sin \theta&= 1\\ \theta&= \dfrac{\pi}{2} \end{align*}\]

Exercise \(\PageIndex{7}\)

Solve the quadratic equation \(2{\cos}^2 \theta+\cos \theta=0\).

\(\dfrac{\pi}{2}, \space \dfrac{2\pi}{3}, \space \dfrac{4\pi}{3}, \space \dfrac{3\pi}{2}\)

Solving Trigonometric Equations Using Fundamental Identities

While algebra can be used to solve a number of trigonometric equations, we can also use the fundamental identities because they make solving equations simpler. Remember that the techniques we use for solving are not the same as those for verifying identities. The basic rules of algebra apply here, as opposed to rewriting one side of the identity to match the other side. In the next example, we use two identities to simplify the equation.

Example \(\PageIndex{8A}\): Use Identities to Solve an Equation

Use identities to solve exactly the trigonometric equation over the interval \(0≤x<2\pi\).

\(\cos x \cos(2x)+\sin x \sin(2x)=\dfrac{\sqrt{3}}{2}\)

Notice that the left side of the equation is the difference formula for cosine.

\[\begin{align*} \cos x \cos(2x)+\sin x \sin(2x)&= \dfrac{\sqrt{3}}{2}\\ \cos(x-2x)&= \dfrac{\sqrt{3}}{2}\qquad \text{Difference formula for cosine}\\ \cos(-x)&= \dfrac{\sqrt{3}}{2}\qquad \text{Use the negative angle identity.}\\ \cos x&= \dfrac{\sqrt{3}}{2} \end{align*}\]

From the unit circle in the section on Sum and Difference Identities , we see that \(\cos x=\dfrac{\sqrt{3}}{2}\) when \(x=\dfrac{\pi}{6},\space \dfrac{11\pi}{6}\).

Example \(\PageIndex{8B}\): Solving the Equation Using a Double-Angle Formula

Solve the equation exactly using a double-angle formula: \(\cos(2\theta)=\cos \theta\).

We have three choices of expressions to substitute for the double-angle of cosine. As it is simpler to solve for one trigonometric function at a time, we will choose the double-angle identity involving only cosine:

\[\begin{align*} \cos(2\theta)&= \cos \theta\\ 2{\cos}^2 \theta-1&= \cos \theta\\ 2 {\cos}^2 \theta-\cos \theta-1&= 0\\ (2 \cos \theta+1)(\cos \theta-1)&= 0\\ 2 \cos \theta+1&= 0\\ \cos \theta&= -\dfrac{1}{2}\\ \cos \theta-1&= 0\\ \cos \theta&= 1 \end{align*}\]

So, if \(\cos \theta=−\dfrac{1}{2}\),then \(\theta=\dfrac{2\pi}{3}\pm 2\pi k\) and \(\theta=\dfrac{4\pi}{3}\pm 2\pi k\); if \(\cos \theta=1\),then \(\theta=0\pm 2\pi k\).

Example \(\PageIndex{8C}\): Solving an Equation Using an Identity

Solve the equation exactly using an identity: \(3 \cos \theta+3=2 {\sin}^2 \theta\), \(0≤\theta<2\pi\).

If we rewrite the right side, we can write the equation in terms of cosine:

\[\begin{align*} 3 \cos \theta+3&= 2 {\sin}^2 \theta\\ 3 \cos \theta+3&= 2(1-{\cos}^2 \theta)\\ 3 \cos \theta+3&= 2-2{\cos}^2 \theta\\ 2 {\cos}^2 \theta+3 \cos \theta+1&= 0\\ (2 \cos \theta+1)(\cos \theta+1)&= 0\\ 2 \cos \theta+1&= 0\\ \cos \theta&= -\dfrac{1}{2}\\ \theta&= \dfrac{2\pi}{3},\space \dfrac{4\pi}{3}\\ \cos \theta+1&= 0\\ \cos \theta&= -1\\ \theta&= \pi\\ \end{align*}\]

Our solutions are \(\theta=\dfrac{2\pi}{3},\space \dfrac{4\pi}{3},\space \pi\).

Solving Trigonometric Equations with Multiple Angles

Sometimes it is not possible to solve a trigonometric equation with identities that have a multiple angle, such as \(\sin(2x)\) or \(\cos(3x)\). When confronted with these equations, recall that \(y=\sin(2x)\) is a horizontal compression by a factor of 2 of the function \(y=\sin x\). On an interval of \(2\pi\),we can graph two periods of \(y=\sin(2x)\),as opposed to one cycle of \(y=\sin x\). This compression of the graph leads us to believe there may be twice as many x -intercepts or solutions to \(\sin(2x)=0\) compared to \(\sin x=0\). This information will help us solve the equation.

Example \(\PageIndex{9}\): Solving a Multiple Angle Trigonometric Equation

Solve exactly: \(\cos(2x)=\dfrac{1}{2}\) on \([ 0,2\pi )\).

We can see that this equation is the standard equation with a multiple of an angle. If \(\cos(\alpha)=\dfrac{1}{2}\),we know \(\alpha\) is in quadrants I and IV. While \(\theta={\cos}^{−1} \dfrac{1}{2}\) will only yield solutions in quadrants I and II, we recognize that the solutions to the equation \(\cos \theta=\dfrac{1}{2}\) will be in quadrants I and IV.

Therefore, the possible angles are \(\theta=\dfrac{\pi}{3}\) and \(\theta=\dfrac{5\pi}{3}\). So, \(2x=\dfrac{\pi}{3}\) or \(2x=\dfrac{5\pi}{3}\), which means that \(x=\dfrac{\pi}{6}\) or \(x=\dfrac{5\pi}{6}\). Does this make sense? Yes, because \(\cos\left(2\left(\dfrac{\pi}{6}\right)\right)=\cos\left(\dfrac{\pi}{3}\right)=\dfrac{1}{2}\).

Are there any other possible answers? Let us return to our first step.

In quadrant I, \(2x=\dfrac{\pi}{3}\), so \(x=\dfrac{\pi}{6}\) as noted. Let us revolve around the circle again:

\[\begin{align*} 2x&= \dfrac{\pi}{3}+2\pi\\ &= \dfrac{\pi}{3}+\dfrac{6\pi}{3}\\ &= \dfrac{7\pi}{3}\\ x&= \dfrac{7\pi}{6}\\ \text {One more rotation yields}\\ 2x&= \dfrac{\pi}{3}+4\pi\\ &= \dfrac{\pi}{3}+\dfrac{12\pi}{3}\\ &= \dfrac{13\pi}{3}\\ \end{align*}\]

\(x=\dfrac{13\pi}{6}>2\pi\), so this value for \(x\) is larger than \(2\pi\), so it is not a solution on \([ 0,2\pi )\).

In quadrant IV, \(2x=\dfrac{5\pi}{3}\), so \(x=\dfrac{5\pi}{6}\) as noted. Let us revolve around the circle again:

\[\begin{align*} 2x&= \dfrac{5\pi}{3}+2\pi\\ &= \dfrac{5\pi}{3}+\dfrac{6\pi}{3}\\ &= \dfrac{11\pi}{3} \end{align*}\]

so \(x=\dfrac{11\pi}{6}\).

One more rotation yields

\[\begin{align*} 2x&= \dfrac{5\pi}{3}+4\pi\\ &= \dfrac{5\pi}{3}+\dfrac{12\pi}{3}\\ &= \dfrac{17\pi}{3} \end{align*}\]

\(x=\dfrac{17\pi}{6}>2\pi\),so this value for \(x\) is larger than \(2\pi\),so it is not a solution on \([ 0,2\pi )\)  .

Our solutions are \(x=\dfrac{\pi}{6}, \space \dfrac{5\pi}{6}, \space \dfrac{7\pi}{6}\), and \(\dfrac{11\pi}{6}\). Note that whenever we solve a problem in the form of \(sin(nx)=c\), we must go around the unit circle \(n\) times.

Solving Right Triangle Problems

We can now use all of the methods we have learned to solve problems that involve applying the properties of right triangles and the Pythagorean Theorem. We begin with the familiar Pythagorean Theorem,

\[a^2+b^2=c^2 \label{Pythagorean}\]

and model an equation to fit a situation.

Example \(\PageIndex{10A}\): Using the Pythagorean Theorem to Model an Equation

One of the cables that anchors the center of the London Eye Ferris wheel to the ground must be replaced. The center of the Ferris wheel is \(69.5\) meters above the ground, and the second anchor on the ground is \(23\) meters from the base of the Ferris wheel. Approximately how long is the cable, and what is the angle of elevation (from ground up to the center of the Ferris wheel)? See Figure \(\PageIndex{4}\).

Basic diagram of a ferris wheel (circle) and its support cables (form a right triangle). One cable runs from the center of the circle to the ground (outside the circle), is perpendicular to the ground, and has length 69.5. Another cable of unknown length (the hypotenuse) runs from the center of the circle to the ground 23 feet away from the other cable at an angle of theta degrees with the ground. So, in closing, there is a right triangle with base 23, height 69.5, hypotenuse unknown, and angle between base and hypotenuse of theta degrees.

Use the Pythagorean Theorem (Equation \ref{Pythagorean}) and the properties of right triangles to model an equation that fits the problem. Using the information given, we can draw a right triangle. We can find the length of the cable with the Pythagorean Theorem.

\[\begin{align*} a^2+b^2&= c^2\\ {(23)}^2+{(69.5)}^2&\approx 5359\\ \sqrt{5359}&\approx 73.2\space m \end{align*}\]

The angle of elevation is \(\theta\),formed by the second anchor on the ground and the cable reaching to the center of the wheel. We can use the tangent function to find its measure. Round to two decimal places.

\[\begin{align*} \tan \theta&= 69.523\\ {\tan}^{-1}(69.523)&\approx 1.2522\\ &\approx 71.69^{\circ} \end{align*}\]

The angle of elevation is approximately \(71.7°\), and the length of the cable is \(73.2\) meters.

Example \(\PageIndex{10B}\): Using the Pythagorean Theorem to Model an Abstract Problem

OSHA safety regulations require that the base of a ladder be placed \(1\) foot from the wall for every \(4\) feet of ladder length. Find the angle that a ladder of any length forms with the ground and the height at which the ladder touches the wall.

For any length of ladder, the base needs to be a distance from the wall equal to one fourth of the ladder’s length. Equivalently, if the base of the ladder is “ a” feet from the wall, the length of the ladder will be \(4a\) feet. See Figure \(\PageIndex{5}\).

Diagram of a right triangle with base length a, height length b, hypotenuse length 4a. Opposite the height is an angle of theta degrees, and opposite the hypotenuse is an angle of 90 degrees.

The side adjacent to \(\theta\) is \(a\) and the hypotenuse is \(4a\). Thus,

\[\begin{align*} \cos \theta&= \dfrac{a}{4a}\\ &= \dfrac{1}{4}\\ {\cos}^{-1}\left (\dfrac{1}{4}\right )&\approx 75.5^{\circ} \end{align*}\]

The elevation of the ladder forms an angle of \(75.5°\) with the ground. The height at which the ladder touches the wall can be found using the Pythagorean Theorem:

\[\begin{align*} a^2+b^2&= {(4a)}^2\\ b^2&= {(4a)}^2-a^2\\ b^2&= 16a^2-a^2\\ b^2&= 15a^2\\ b&= a\sqrt{15} \end{align*}\]

Thus, the ladder touches the wall at \(a\sqrt{15}\) feet from the ground.

Access these online resources for additional instruction and practice with solving trigonometric equations.

  • Solving Trigonometric Equations I
  • Solving Trigonometric Equations II
  • Solving Trigonometric Equations III
  • Solving Trigonometric Equations IV
  • Solving Trigonometric Equations V
  • Solving Trigonometric Equations VI

Key Concepts

  • When solving linear trigonometric equations, we can use algebraic techniques just as we do solving algebraic equations. Look for patterns, like the difference of squares, quadratic form, or an expression that lends itself well to substitution. See Example \(\PageIndex{1}\), Example \(\PageIndex{2}\), and Example \(\PageIndex{3}\).
  • Equations involving a single trigonometric function can be solved or verified using the unit circle. See Example \(\PageIndex{4}\), Example \(\PageIndex{5}\), and Example \(\PageIndex{6}\), and Example \(\PageIndex{7}\).
  • We can also solve trigonometric equations using a graphing calculator. See Example \(\PageIndex{8}\) and Example \(\PageIndex{9}\).
  • Many equations appear quadratic in form. We can use substitution to make the equation appear simpler, and then use the same techniques we use solving an algebraic quadratic: factoring, the quadratic formula, etc. See Example \(\PageIndex{10}\), Example \(\PageIndex{11}\), Example \(\PageIndex{12}\), and Example \(\PageIndex{13}\).
  • We can also use the identities to solve trigonometric equation. See Example \(\PageIndex{14}\), Example \(\PageIndex{15}\), and Example \(\PageIndex{16}\).
  • We can use substitution to solve a multiple-angle trigonometric equation, which is a compression of a standard trigonometric function. We will need to take the compression into account and verify that we have found all solutions on the given interval. See Example \(\PageIndex{17}\).
  • Real-world scenarios can be modeled and solved using the Pythagorean Theorem and trigonometric functions. See Example \(\PageIndex{18}\).

Trigonometric Equation Calculator

Instructions: Use calculator to solve trigonometric equations you provide, showing all the steps. Please type in the trigonometric equation you want to carry out in the box below.

solving trig equations with domain calculator

ABout this Trigonometric Equation Calculator

This calculator will allow you to solve trig equations, showing all the steps of the way. All you need to do is to provide a valid trigonometric equation, with an unknown (x). It could be something simple as 'sin(x) = 1/2', or something more complex like 'sin^2(x) = cos(x) + tan(x)'.

Once you are done typing your equation, just go ahead and click on "Solve" to get all the details of the processes of finding the solutions, if solutions can be found.

Trigonometric properties and rules almost always allow to reduce most trig equations into simpler ones, so this type of equation is one type that often times lead to solutions, but it can be extremely cumbersome at times.

Trigonometric Equation Calculator

What is a trigonometric Equation?

A trigonometric equation, in the simplest possible terms, is a math equation where the unknown x is inside of a trigonometric expression.

For example, the following expression is a trigonometric eqn:

Why? Simply because x appears inside of trig expression sine. Or for example:

Now, these two are trig eqns, but the difference between the two is that for the first one, x appears ONLY inside of sine, whereas in the second one x appears inside of a trig function (tangent), but it also appears outside. This will usually make it difficult (or impossible) to solve the equation.

How to solve trigonometric equations

  • Step 1: Make sure you are dealing with a trigonometric equation. Non-trigonometric equations will likely require a different approach
  • Step 2: Make sure that the unknown x is inside of the trigonometric expression , but x does not appear outside a trig expression. If that is the case, it is likely you won't be able to solve the equation with elementary methods
  • Step 3: Conduct an appropriate substitution, by first expressing all the trig functions present in the equation into one type (typically sine), and then use a substitution involving sine
  • Step 4: With a little luck and if you did the correct substitution, you have reduced the original trigonometric equation into a polynomial equation to solve .

One of the key trig rules you need to use the ability to express all trigonometric functions in terms of any fixed trigonometric function. For example, we can write cosine in terms of sine:

Trigonometric Substitutions

Using trigonometric identities and substitutions is your way to go in this case. For example, suppose you want to solve this:

So we know this is a trig equation, and we know we can write cosine in terms of sine, so we do this:

Now what? Well, we can use the substitution: \(u = \sin x\), so the equation above becomes:

which is a rational equation , which by using simple algebraic manipulation means that we need to solve a polynomial equation in order to solve the original trig equation.

 Trigonometric Equation

Application of trigonometry

  • Step 1: All things mechanical: In manufacturing mechanical parts, circles and trigonometry play a crucial role
  • Step 2: Analysis of periodic functions: Many phenomena are tightly related to periodicity, the point at which trigonometry comes to play
  • Step 3: Advanced math: Mathematicians love their Fourier Series and Transform, which play a crucial role in spectral analysis

Circles and all their symmetry are so really important in real life, and trigonometry is the language by which we can quantify circles and its relationships. Solving trigonometric equations is at the center of math.

Why would you solve trigonometric equations

Trigonometric equations carry lots of value in practice especially in Engineering. Notable properties such as the period and frequency open a full spectrum of applications.

Circular structures play a crucial role in everything mechanical that we use today. Circles are synonymous of trigonometry, and trigonometric equations are at the center of it.

 Trigonometric Calculator

Example: Solving simple trig equations

Solve: \(\sin(x) = \frac{1}{2}\)

We need to solve the following given trigonometric equation equation:

The following is obtained:

By direct application of the properties of the inverse trigonometric function \( \arcsin(\cdot)\), as well as the properties of the trigonometric function \( \sin\left(x\right)\), we obtain that

Therefore, solving for \(x\) for the given equation leads to the solutions \(x=\frac{5}{6}\pi{}+2\pi{}K_1,\,\,x=\frac{1}{6}\pi{}+2\pi{}K_2\), for \(K_1, K_2\) arbitrary integer constants.

More equation calculators

Our trigonometric equation with steps will come in handy when dealing with equations with specific structure. If you are unsure of the type of equation you are dealing with, you can use our general equation solver , which will figure out the structure of the given equation, and will find a suitable approach.

The main difficulty with solving equations that are not linear equation or polynomial equation is that there is not a specific route to follow, nor there is any guarantee you will find solutions.

Usually, the strategy consists of simplifying expressions as much as possible, and after doing that, it is usually nowhere land, where you need to try whatever feels suitable.

Naturally, the idea is to try to reduce the equation to a simpler equation, by using some kind of substitution and a multi-step process, where you first find solutions of an auxiliary solution, which gives you CANDIDATES to the original equation. You would like to solve a linear equation , or even a quadratic equation , but perhaps the reduction you get will be a bit less generous.

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  • 7.5 Solving Trigonometric Equations
  • Introduction to Functions
  • 1.1 Functions and Function Notation
  • 1.2 Domain and Range
  • 1.3 Rates of Change and Behavior of Graphs
  • 1.4 Composition of Functions
  • 1.5 Transformation of Functions
  • 1.6 Absolute Value Functions
  • 1.7 Inverse Functions
  • Key Equations
  • Key Concepts
  • Review Exercises
  • Practice Test
  • Introduction to Linear Functions
  • 2.1 Linear Functions
  • 2.2 Graphs of Linear Functions
  • 2.3 Modeling with Linear Functions
  • 2.4 Fitting Linear Models to Data
  • Introduction to Polynomial and Rational Functions
  • 3.1 Complex Numbers
  • 3.2 Quadratic Functions
  • 3.3 Power Functions and Polynomial Functions
  • 3.4 Graphs of Polynomial Functions
  • 3.5 Dividing Polynomials
  • 3.6 Zeros of Polynomial Functions
  • 3.7 Rational Functions
  • 3.8 Inverses and Radical Functions
  • 3.9 Modeling Using Variation
  • Introduction to Exponential and Logarithmic Functions
  • 4.1 Exponential Functions
  • 4.2 Graphs of Exponential Functions
  • 4.3 Logarithmic Functions
  • 4.4 Graphs of Logarithmic Functions
  • 4.5 Logarithmic Properties
  • 4.6 Exponential and Logarithmic Equations
  • 4.7 Exponential and Logarithmic Models
  • 4.8 Fitting Exponential Models to Data
  • Introduction to Trigonometric Functions
  • 5.2 Unit Circle: Sine and Cosine Functions
  • 5.3 The Other Trigonometric Functions
  • 5.4 Right Triangle Trigonometry
  • Introduction to Periodic Functions
  • 6.1 Graphs of the Sine and Cosine Functions
  • 6.2 Graphs of the Other Trigonometric Functions
  • 6.3 Inverse Trigonometric Functions
  • Introduction to Trigonometric Identities and Equations
  • 7.1 Solving Trigonometric Equations with Identities
  • 7.2 Sum and Difference Identities
  • 7.3 Double-Angle, Half-Angle, and Reduction Formulas
  • 7.4 Sum-to-Product and Product-to-Sum Formulas
  • 7.6 Modeling with Trigonometric Functions
  • Introduction to Further Applications of Trigonometry
  • 8.1 Non-right Triangles: Law of Sines
  • 8.2 Non-right Triangles: Law of Cosines
  • 8.3 Polar Coordinates
  • 8.4 Polar Coordinates: Graphs
  • 8.5 Polar Form of Complex Numbers
  • 8.6 Parametric Equations
  • 8.7 Parametric Equations: Graphs
  • 8.8 Vectors
  • Introduction to Systems of Equations and Inequalities
  • 9.1 Systems of Linear Equations: Two Variables
  • 9.2 Systems of Linear Equations: Three Variables
  • 9.3 Systems of Nonlinear Equations and Inequalities: Two Variables
  • 9.4 Partial Fractions
  • 9.5 Matrices and Matrix Operations
  • 9.6 Solving Systems with Gaussian Elimination
  • 9.7 Solving Systems with Inverses
  • 9.8 Solving Systems with Cramer's Rule
  • Introduction to Analytic Geometry
  • 10.1 The Ellipse
  • 10.2 The Hyperbola
  • 10.3 The Parabola
  • 10.4 Rotation of Axes
  • 10.5 Conic Sections in Polar Coordinates
  • Introduction to Sequences, Probability and Counting Theory
  • 11.1 Sequences and Their Notations
  • 11.2 Arithmetic Sequences
  • 11.3 Geometric Sequences
  • 11.4 Series and Their Notations
  • 11.5 Counting Principles
  • 11.6 Binomial Theorem
  • 11.7 Probability
  • Introduction to Calculus
  • 12.1 Finding Limits: Numerical and Graphical Approaches
  • 12.2 Finding Limits: Properties of Limits
  • 12.3 Continuity
  • 12.4 Derivatives
  • A | Basic Functions and Identities

Learning Objectives

In this section, you will:

  • Solve linear trigonometric equations in sine and cosine.
  • Solve equations involving a single trigonometric function.
  • Solve trigonometric equations using a calculator.
  • Solve trigonometric equations that are quadratic in form.
  • Solve trigonometric equations using fundamental identities.
  • Solve trigonometric equations with multiple angles.
  • Solve right triangle problems.

Thales of Miletus (circa 625–547 BC) is known as the founder of geometry. The legend is that he calculated the height of the Great Pyramid of Giza in Egypt using the theory of similar triangles , which he developed by measuring the shadow of his staff. Based on proportions, this theory has applications in a number of areas, including fractal geometry, engineering, and architecture. Often, the angle of elevation and the angle of depression are found using similar triangles.

In earlier sections of this chapter, we looked at trigonometric identities. Identities are true for all values in the domain of the variable. In this section, we begin our study of trigonometric equations to study real-world scenarios such as the finding the dimensions of the pyramids.

Solving Linear Trigonometric Equations in Sine and Cosine

Trigonometric equations are, as the name implies, equations that involve trigonometric functions. Similar in many ways to solving polynomial equations or rational equations, only specific values of the variable will be solutions, if there are solutions at all. Often we will solve a trigonometric equation over a specified interval. However, just as often, we will be asked to find all possible solutions, and as trigonometric functions are periodic, solutions are repeated within each period. In other words, trigonometric equations may have an infinite number of solutions. Additionally, like rational equations, the domain of the function must be considered before we assume that any solution is valid. The period of both the sine function and the cosine function is 2 π . 2 π . In other words, every 2 π 2 π units, the y- values repeat. If we need to find all possible solutions, then we must add 2 π k , 2 π k , where k k is an integer, to the initial solution. Recall the rule that gives the format for stating all possible solutions for a function where the period is 2 π : 2 π :

There are similar rules for indicating all possible solutions for the other trigonometric functions. Solving trigonometric equations requires the same techniques as solving algebraic equations. We read the equation from left to right, horizontally, like a sentence. We look for known patterns, factor, find common denominators, and substitute certain expressions with a variable to make solving a more straightforward process. However, with trigonometric equations, we also have the advantage of using the identities we developed in the previous sections.

Solving a Linear Trigonometric Equation Involving the Cosine Function

Find all possible exact solutions for the equation cos θ = 1 2 . cos θ = 1 2 .

From the unit circle , we know that

These are the solutions in the interval [ 0 , 2 π ] . [ 0 , 2 π ] . All possible solutions are given by

where k k is an integer.

Solving a Linear Equation Involving the Sine Function

Find all possible exact solutions for the equation sin t = 1 2 . sin t = 1 2 .

Solving for all possible values of t means that solutions include angles beyond the period of 2 π . 2 π . From Figure 2 , we can see that the solutions are π 6 π 6 and 5 π 6 . 5 π 6 . But the problem is asking for all possible values that solve the equation. Therefore, the answer is

Given a trigonometric equation, solve using algebra .

  • Look for a pattern that suggests an algebraic property, such as the difference of squares or a factoring opportunity.
  • Substitute the trigonometric expression with a single variable, such as x x or u . u .
  • Solve the equation the same way an algebraic equation would be solved.
  • Substitute the trigonometric expression back in for the variable in the resulting expressions.
  • Solve for the angle.

Solve the Trigonometric Equation in Linear Form

Solve the equation exactly: 2 cos θ − 3 = − 5 , 0 ≤ θ < 2 π . 2 cos θ − 3 = − 5 , 0 ≤ θ < 2 π .

Use algebraic techniques to solve the equation.

Solve exactly the following linear equation on the interval [ 0 , 2 π ) : 2 sin x + 1 = 0. [ 0 , 2 π ) : 2 sin x + 1 = 0.

Solving Equations Involving a Single Trigonometric Function

When we are given equations that involve only one of the six trigonometric functions, their solutions involve using algebraic techniques and the unit circle (see Figure 2 ). We need to make several considerations when the equation involves trigonometric functions other than sine and cosine. Problems involving the reciprocals of the primary trigonometric functions need to be viewed from an algebraic perspective. In other words, we will write the reciprocal function, and solve for the angles using the function. Also, an equation involving the tangent function is slightly different from one containing a sine or cosine function. First, as we know, the period of tangent is π , π , not 2 π . 2 π . Further, the domain of tangent is all real numbers with the exception of odd integer multiples of π 2 , π 2 , unless, of course, a problem places its own restrictions on the domain.

Solving a Problem Involving a Single Trigonometric Function

Solve the problem exactly: 2 sin 2 θ − 1 = 0 , 0 ≤ θ < 2 π . 2 sin 2 θ − 1 = 0 , 0 ≤ θ < 2 π .

As this problem is not easily factored, we will solve using the square root property. First, we use algebra to isolate sin θ . sin θ . Then we will find the angles.

Solving a Trigonometric Equation Involving Cosecant

Solve the following equation exactly: csc θ = − 2 , 0 ≤ θ < 4 π . csc θ = − 2 , 0 ≤ θ < 4 π .

We want all values of θ θ for which csc θ = − 2 csc θ = − 2 over the interval 0 ≤ θ < 4 π . 0 ≤ θ < 4 π .

As sin θ = − 1 2 , sin θ = − 1 2 , notice that all four solutions are in the third and fourth quadrants.

Solving an Equation Involving Tangent

Solve the equation exactly: tan ( θ − π 2 ) = 1 , 0 ≤ θ < 2 π . tan ( θ − π 2 ) = 1 , 0 ≤ θ < 2 π .

Recall that the tangent function has a period of π . π . On the interval [ 0 , π ) , [ 0 , π ) , and at the angle of π 4 , π 4 , the tangent has a value of 1. However, the angle we want is ( θ − π 2 ) . ( θ − π 2 ) . Thus, if tan ( π 4 ) = 1 , tan ( π 4 ) = 1 , then

Over the interval [ 0 , 2 π ) , [ 0 , 2 π ) , we have two solutions:

Find all solutions for tan x = 3 . tan x = 3 .

Identify all Solutions to the Equation Involving Tangent

Identify all exact solutions to the equation 2 ( tan x + 3 ) = 5 + tan x , 0 ≤ x < 2 π . 2 ( tan x + 3 ) = 5 + tan x , 0 ≤ x < 2 π .

We can solve this equation using only algebra. Isolate the expression tan x tan x on the left side of the equals sign.

There are two angles on the unit circle that have a tangent value of −1 : θ = 3 π 4 −1 : θ = 3 π 4 and θ = 7 π 4 . θ = 7 π 4 .

Solve Trigonometric Equations Using a Calculator

Not all functions can be solved exactly using only the unit circle. When we must solve an equation involving an angle other than one of the special angles, we will need to use a calculator. Make sure it is set to the proper mode, either degrees or radians, depending on the criteria of the given problem.

Using a Calculator to Solve a Trigonometric Equation Involving Sine

Use a calculator to solve the equation sin θ = 0.8 , sin θ = 0.8 , where θ θ is in radians.

Make sure mode is set to radians. To find θ , θ , use the inverse sine function. On most calculators, you will need to push the 2 ND button and then the SIN button to bring up the sin − 1 sin − 1 function. What is shown on the screen is sin − 1 ( . sin − 1 ( . The calculator is ready for the input within the parentheses. For this problem, we enter sin − 1 ( 0.8 ) , sin − 1 ( 0.8 ) , and press ENTER. Thus, to four decimals places,

The solution is

The angle measurement in degrees is

Note that a calculator will only return an angle in quadrants I or IV for the sine function, since that is the range of the inverse sine. The other angle is obtained by using π − θ . π − θ .

Using a Calculator to Solve a Trigonometric Equation Involving Secant

Use a calculator to solve the equation sec θ = −4 , sec θ = −4 , giving your answer in radians.

We can begin with some algebra.

Check that the MODE is in radians. Now use the inverse cosine function.

Since π 2 ≈ 1.57 π 2 ≈ 1.57 and π ≈ 3.14 , π ≈ 3.14 , 1.8235 is between these two numbers, thus θ ≈ 1 .8235 θ ≈ 1 .8235 is in quadrant II. Cosine is also negative in quadrant III. Note that a calculator will only return an angle in quadrants I or II for the cosine function, since that is the range of the inverse cosine. See Figure 2 .

So, we also need to find the measure of the angle in quadrant III. In quadrant III, the reference angle is θ ' ≈ π − 1 .8235 ≈ 1 .3181 . θ ' ≈ π − 1 .8235 ≈ 1 .3181 . The other solution in quadrant III is π + 1 .3181 ≈ 4 .4597 . π + 1 .3181 ≈ 4 .4597 .

The solutions are 1.8235 ± 2 π k 1.8235 ± 2 π k and 4.4597 ± 2 π k . 4.4597 ± 2 π k .

Solve cos θ = − 0.2. cos θ = − 0.2.

Solving Trigonometric Equations in Quadratic Form

Solving a quadratic equation may be more complicated, but once again, we can use algebra as we would for any quadratic equation. Look at the pattern of the equation. Is there more than one trigonometric function in the equation, or is there only one? Which trigonometric function is squared? If there is only one function represented and one of the terms is squared, think about the standard form of a quadratic. Replace the trigonometric function with a variable such as x x or u . u . If substitution makes the equation look like a quadratic equation, then we can use the same methods for solving quadratics to solve the trigonometric equations.

Solving a Trigonometric Equation in Quadratic Form

Solve the equation exactly: cos 2 θ + 3 cos θ − 1 = 0 , 0 ≤ θ < 2 π . cos 2 θ + 3 cos θ − 1 = 0 , 0 ≤ θ < 2 π .

We begin by using substitution and replacing cos θ θ with x . x . It is not necessary to use substitution, but it may make the problem easier to solve visually. Let cos θ = x . cos θ = x . We have

The equation cannot be factored, so we will use the quadratic formula x = − b ± b 2 − 4 a c 2 a . x = − b ± b 2 − 4 a c 2 a .

Replace x x with cos θ , cos θ , and solve. Thus,

Note that only the + sign is used. This is because we get an error when we solve θ = cos − 1 ( − 3 − 13 2 ) θ = cos − 1 ( − 3 − 13 2 ) on a calculator, since the domain of the inverse cosine function is [ − 1 , 1 ] . [ − 1 , 1 ] . However, there is a second solution:

This terminal side of the angle lies in quadrant I. Since cosine is also positive in quadrant IV, the second solution is

Solving a Trigonometric Equation in Quadratic Form by Factoring

Solve the equation exactly: 2 sin 2 θ − 5 sin θ + 3 = 0 , 0 ≤ θ ≤ 2 π . 2 sin 2 θ − 5 sin θ + 3 = 0 , 0 ≤ θ ≤ 2 π .

Using grouping, this quadratic can be factored. Either make the real substitution, sin θ = u , sin θ = u , or imagine it, as we factor:

Now set each factor equal to zero.

Next solve for θ : sin θ ≠ 3 2 , θ : sin θ ≠ 3 2 , as the range of the sine function is [ −1 , 1 ] . [ −1 , 1 ] . However, sin θ = 1 , sin θ = 1 , giving the solution π 2 . π 2 .

Make sure to check all solutions on the given domain as some factors have no solution.

Solve sin 2 θ = 2 cos θ + 2 , 0 ≤ θ ≤ 2 π . sin 2 θ = 2 cos θ + 2 , 0 ≤ θ ≤ 2 π . [Hint: Make a substitution to express the equation only in terms of cosine.]

Solving a Trigonometric Equation Using Algebra

Solve exactly:

This problem should appear familiar as it is similar to a quadratic. Let sin θ = x . sin θ = x . The equation becomes 2 x 2 + x = 0. 2 x 2 + x = 0. We begin by factoring:

Set each factor equal to zero.

Then, substitute back into the equation the original expression sin θ sin θ for x . x . Thus,

The solutions within the domain 0 ≤ θ < 2 π 0 ≤ θ < 2 π are 0 , π , 7 π 6 , 11 π 6 . 0 , π , 7 π 6 , 11 π 6 .

If we prefer not to substitute, we can solve the equation by following the same pattern of factoring and setting each factor equal to zero.

We can see the solutions on the graph in Figure 3 . On the interval 0 ≤ θ < 2 π , 0 ≤ θ < 2 π , the graph crosses the x- axis four times, at the solutions noted. Notice that trigonometric equations that are in quadratic form can yield up to four solutions instead of the expected two that are found with quadratic equations. In this example, each solution (angle) corresponding to a positive sine value will yield two angles that would result in that value.

We can verify the solutions on the unit circle in Figure 2 as well.

Solving a Trigonometric Equation Quadratic in Form

Solve the equation quadratic in form exactly: 2 sin 2 θ − 3 sin θ + 1 = 0 , 0 ≤ θ < 2 π . 2 sin 2 θ − 3 sin θ + 1 = 0 , 0 ≤ θ < 2 π .

We can factor using grouping. Solution values of θ θ can be found on the unit circle:

Solve the quadratic equation 2 cos 2 θ + cos θ = 0. 2 cos 2 θ + cos θ = 0.

Solving Trigonometric Equations Using Fundamental Identities

While algebra can be used to solve a number of trigonometric equations, we can also use the fundamental identities because they make solving equations simpler. Remember that the techniques we use for solving are not the same as those for verifying identities. The basic rules of algebra apply here, as opposed to rewriting one side of the identity to match the other side. In the next example, we use two identities to simplify the equation.

Use Identities to Solve an Equation

Use identities to solve exactly the trigonometric equation over the interval 0 ≤ x < 2 π . 0 ≤ x < 2 π .

Notice that the left side of the equation is the difference formula for cosine.

From the unit circle in Figure 2 , we see that cos x = 3 2 cos x = 3 2 when x = π 6 , 11 π 6 . x = π 6 , 11 π 6 .

Solving the Equation Using a Double-Angle Formula

Solve the equation exactly using a double-angle formula: cos ( 2 θ ) = cos θ . cos ( 2 θ ) = cos θ .

We have three choices of expressions to substitute for the double-angle of cosine. As it is simpler to solve for one trigonometric function at a time, we will choose the double-angle identity involving only cosine:

So, if cos θ = − 1 2 , cos θ = − 1 2 , then θ = 2 π 3 ± 2 π k θ = 2 π 3 ± 2 π k and θ = 4 π 3 ± 2 π k ; θ = 4 π 3 ± 2 π k ; if cos θ = 1 , cos θ = 1 , then θ = 0 ± 2 π k . θ = 0 ± 2 π k .

Solving an Equation Using an Identity

Solve the equation exactly using an identity: 3 cos θ + 3 = 2 sin 2 θ , 0 ≤ θ < 2 π . 3 cos θ + 3 = 2 sin 2 θ , 0 ≤ θ < 2 π .

If we rewrite the right side, we can write the equation in terms of cosine:

Our solutions are 2 π 3 , 4 π 3 , π . 2 π 3 , 4 π 3 , π .

Solving Trigonometric Equations with Multiple Angles

Sometimes it is not possible to solve a trigonometric equation with identities that have a multiple angle, such as sin ( 2 x ) sin ( 2 x ) or cos ( 3 x ) . cos ( 3 x ) . When confronted with these equations, recall that y = sin ( 2 x ) y = sin ( 2 x ) is a horizontal compression by a factor of 2 of the function y = sin x . y = sin x . On an interval of 2 π , 2 π , we can graph two periods of y = sin ( 2 x ) , y = sin ( 2 x ) , as opposed to one cycle of y = sin x . y = sin x . This compression of the graph leads us to believe there may be twice as many x -intercepts or solutions to sin ( 2 x ) = 0 sin ( 2 x ) = 0 compared to sin x = 0. sin x = 0. This information will help us solve the equation.

Solving a Multiple Angle Trigonometric Equation

Solve exactly: cos ( 2 x ) = 1 2 cos ( 2 x ) = 1 2 on [ 0 , 2 π ) . [ 0 , 2 π ) .

We can see that this equation is the standard equation with a multiple of an angle. If cos ( α ) = 1 2 , cos ( α ) = 1 2 , we know α α is in quadrants I and IV. While θ = cos − 1 1 2 θ = cos − 1 1 2 will only yield solutions in quadrants I and II, we recognize that the solutions to the equation cos θ = 1 2 cos θ = 1 2 will be in quadrants I and IV.

Therefore, the possible angles are θ = π 3 θ = π 3 and θ = 5 π 3 . θ = 5 π 3 . So, 2 x = π 3 2 x = π 3 or 2 x = 5 π 3 , 2 x = 5 π 3 , which means that x = π 6 x = π 6 or x = 5 π 6 . x = 5 π 6 . Does this make sense? Yes, because cos ( 2 ( π 6 ) ) = cos ( π 3 ) = 1 2 . cos ( 2 ( π 6 ) ) = cos ( π 3 ) = 1 2 .

Are there any other possible answers? Let us return to our first step.

In quadrant I, 2 x = π 3 , 2 x = π 3 , so x = π 6 x = π 6 as noted. Let us revolve around the circle again:

so x = 7 π 6 . x = 7 π 6 .

One more rotation yields

x = 13 π 6 > 2 π , x = 13 π 6 > 2 π , so this value for x x is larger than 2 π , 2 π , so it is not a solution on [ 0 , 2 π ) . [ 0 , 2 π ) .

In quadrant IV, 2 x = 5 π 3 , 2 x = 5 π 3 , so x = 5 π 6 x = 5 π 6 as noted. Let us revolve around the circle again:

so x = 11 π 6 . x = 11 π 6 .

x = 17 π 6 > 2 π , x = 17 π 6 > 2 π , so this value for x x is larger than 2 π , 2 π , so it is not a solution on [ 0 , 2 π ) . [ 0 , 2 π ) .

Our solutions are π 6 , 5 π 6 , 7 π 6 , and  11 π 6 . π 6 , 5 π 6 , 7 π 6 , and  11 π 6 . Note that whenever we solve a problem in the form of sin ( n x ) = c , sin ( n x ) = c , we must go around the unit circle n n times.

Solving Right Triangle Problems

We can now use all of the methods we have learned to solve problems that involve applying the properties of right triangles and the Pythagorean Theorem . We begin with the familiar Pythagorean Theorem, a 2 + b 2 = c 2 , a 2 + b 2 = c 2 , and model an equation to fit a situation.

Using the Pythagorean Theorem to Model an Equation

Use the Pythagorean Theorem, and the properties of right triangles to model an equation that fits the problem.

One of the cables that anchors the center of the London Eye Ferris wheel to the ground must be replaced. The center of the Ferris wheel is 69.5 meters above the ground, and the second anchor on the ground is 23 meters from the base of the Ferris wheel. Approximately how long is the cable, and what is the angle of elevation (from ground up to the center of the Ferris wheel)? See Figure 4 .

Using the information given, we can draw a right triangle. We can find the length of the cable with the Pythagorean Theorem.

The angle of elevation is θ , θ , formed by the second anchor on the ground and the cable reaching to the center of the wheel. We can use the tangent function to find its measure. Round to two decimal places.

The angle of elevation is approximately 71.7 ∘ , 71.7 ∘ , and the length of the cable is 73.2 meters.

Using the Pythagorean Theorem to Model an Abstract Problem

OSHA safety regulations require that the base of a ladder be placed 1 foot from the wall for every 4 feet of ladder length. Find the angle that a ladder of any length forms with the ground and the height at which the ladder touches the wall.

For any length of ladder, the base needs to be a distance from the wall equal to one fourth of the ladder’s length. Equivalently, if the base of the ladder is “ a” feet from the wall, the length of the ladder will be 4 a feet. See Figure 5 .

The side adjacent to θ θ is a and the hypotenuse is 4 a . 4 a . Thus,

The elevation of the ladder forms an angle of 75.5 ∘ 75.5 ∘ with the ground. The height at which the ladder touches the wall can be found using the Pythagorean Theorem:

Thus, the ladder touches the wall at 15 a 15 a feet from the ground.

Access these online resources for additional instruction and practice with solving trigonometric equations.

  • Solving Trigonometric Equations I
  • Solving Trigonometric Equations II
  • Solving Trigonometric Equations III
  • Solving Trigonometric Equations IV
  • Solving Trigonometric Equations V
  • Solving Trigonometric Equations VI

7.5 Section Exercises

Will there always be solutions to trigonometric function equations? If not, describe an equation that would not have a solution. Explain why or why not.

When solving a trigonometric equation involving more than one trig function, do we always want to try to rewrite the equation so it is expressed in terms of one trigonometric function? Why or why not?

When solving linear trig equations in terms of only sine or cosine, how do we know whether there will be solutions?

For the following exercises, find all solutions exactly on the interval 0 ≤ θ < 2 π . 0 ≤ θ < 2 π .

2 sin θ = − 2 2 sin θ = − 2

2 sin θ = 3 2 sin θ = 3

2 cos θ = 1 2 cos θ = 1

2 cos θ = − 2 2 cos θ = − 2

tan θ = −1 tan θ = −1

tan x = 1 tan x = 1

cot x + 1 = 0 cot x + 1 = 0

4 sin 2 x − 2 = 0 4 sin 2 x − 2 = 0

csc 2 x − 4 = 0 csc 2 x − 4 = 0

For the following exercises, solve exactly on [ 0 , 2 π ) . [ 0 , 2 π ) .

2 cos θ = 2 2 cos θ = 2

2 cos θ = −1 2 cos θ = −1

2 sin θ = −1 2 sin θ = −1

2 sin θ = − 3 2 sin θ = − 3

2 sin ( 3 θ ) = 1 2 sin ( 3 θ ) = 1

2 sin ( 2 θ ) = 3 2 sin ( 2 θ ) = 3

2 cos ( 3 θ ) = − 2 2 cos ( 3 θ ) = − 2

cos ( 2 θ ) = − 3 2 cos ( 2 θ ) = − 3 2

2 sin ( π θ ) = 1 2 sin ( π θ ) = 1

2 cos ( π 5 θ ) = 3 2 cos ( π 5 θ ) = 3

For the following exercises, find all exact solutions on [ 0 , 2 π ) . [ 0 , 2 π ) .

sec ( x ) sin ( x ) − 2 sin ( x ) = 0 sec ( x ) sin ( x ) − 2 sin ( x ) = 0

tan ( x ) − 2 sin ( x ) tan ( x ) = 0 tan ( x ) − 2 sin ( x ) tan ( x ) = 0

2 cos 2 t + cos ( t ) = 1 2 cos 2 t + cos ( t ) = 1

2 tan 2 ( t ) = 3 sec ( t ) 2 tan 2 ( t ) = 3 sec ( t )

2 sin ( x ) cos ( x ) − sin ( x ) + 2 cos ( x ) − 1 = 0 2 sin ( x ) cos ( x ) − sin ( x ) + 2 cos ( x ) − 1 = 0

cos 2 θ = 1 2 cos 2 θ = 1 2

sec 2 x = 1 sec 2 x = 1

tan 2 ( x ) = − 1 + 2 tan ( − x ) tan 2 ( x ) = − 1 + 2 tan ( − x )

8 sin 2 ( x ) + 6 sin ( x ) + 1 = 0 8 sin 2 ( x ) + 6 sin ( x ) + 1 = 0

tan 5 ( x ) = tan ( x ) tan 5 ( x ) = tan ( x )

For the following exercises, solve with the methods shown in this section exactly on the interval [ 0 , 2 π ) . [ 0 , 2 π ) .

sin ( 3 x ) cos ( 6 x ) − cos ( 3 x ) sin ( 6 x ) = −0.9 sin ( 3 x ) cos ( 6 x ) − cos ( 3 x ) sin ( 6 x ) = −0.9

sin ( 6 x ) cos ( 11 x ) − cos ( 6 x ) sin ( 11 x ) = −0.1 sin ( 6 x ) cos ( 11 x ) − cos ( 6 x ) sin ( 11 x ) = −0.1

cos ( 2 x ) cos x + sin ( 2 x ) sin x = 1 cos ( 2 x ) cos x + sin ( 2 x ) sin x = 1

6 sin ( 2 t ) + 9 sin t = 0 6 sin ( 2 t ) + 9 sin t = 0

9 cos ( 2 θ ) = 9 cos 2 θ − 4 9 cos ( 2 θ ) = 9 cos 2 θ − 4

sin ( 2 t ) = cos t sin ( 2 t ) = cos t

cos ( 2 t ) = sin t cos ( 2 t ) = sin t

cos ( 6 x ) − cos ( 3 x ) = 0 cos ( 6 x ) − cos ( 3 x ) = 0

For the following exercises, solve exactly on the interval [ 0 , 2 π ) . [ 0 , 2 π ) . Use the quadratic formula if the equations do not factor.

tan 2 x − 3 tan x = 0 tan 2 x − 3 tan x = 0

sin 2 x + sin x − 2 = 0 sin 2 x + sin x − 2 = 0

sin 2 x − 2 sin x − 4 = 0 sin 2 x − 2 sin x − 4 = 0

5 cos 2 x + 3 cos x − 1 = 0 5 cos 2 x + 3 cos x − 1 = 0

3 cos 2 x − 2 cos x − 2 = 0 3 cos 2 x − 2 cos x − 2 = 0

5 sin 2 x + 2 sin x − 1 = 0 5 sin 2 x + 2 sin x − 1 = 0

tan 2 x + 5 tan x − 1 = 0 tan 2 x + 5 tan x − 1 = 0

cot 2 x = − cot x cot 2 x = − cot x

− tan 2 x − tan x − 2 = 0 − tan 2 x − tan x − 2 = 0

For the following exercises, find exact solutions on the interval [ 0 , 2 π ) . [ 0 , 2 π ) . Look for opportunities to use trigonometric identities.

sin 2 x − cos 2 x − sin x = 0 sin 2 x − cos 2 x − sin x = 0

sin 2 x + cos 2 x = 0 sin 2 x + cos 2 x = 0

sin ( 2 x ) − sin x = 0 sin ( 2 x ) − sin x = 0

cos ( 2 x ) − cos x = 0 cos ( 2 x ) − cos x = 0

2 tan x 2 − sec 2 x − sin 2 x = cos 2 x 2 tan x 2 − sec 2 x − sin 2 x = cos 2 x

1 − cos ( 2 x ) = 1 + cos ( 2 x ) 1 − cos ( 2 x ) = 1 + cos ( 2 x )

sec 2 x = 7 sec 2 x = 7

10 sin x cos x = 6 cos x 10 sin x cos x = 6 cos x

−3 sin t = 15 cos t sin t −3 sin t = 15 cos t sin t

4 cos 2 x − 4 = 15 cos x 4 cos 2 x − 4 = 15 cos x

8 sin 2 x + 6 sin x + 1 = 0 8 sin 2 x + 6 sin x + 1 = 0

8 cos 2 θ = 3 − 2 cos θ 8 cos 2 θ = 3 − 2 cos θ

6 cos 2 x + 7 sin x − 8 = 0 6 cos 2 x + 7 sin x − 8 = 0

12 sin 2 t + cos t − 6 = 0 12 sin 2 t + cos t − 6 = 0

tan x = 3 sin x tan x = 3 sin x

cos 3 t = cos t cos 3 t = cos t

For the following exercises, algebraically determine all solutions of the trigonometric equation exactly, then verify the results by graphing the equation and finding the zeros.

6 sin 2 x − 5 sin x + 1 = 0 6 sin 2 x − 5 sin x + 1 = 0

8 cos 2 x − 2 cos x − 1 = 0 8 cos 2 x − 2 cos x − 1 = 0

100 tan 2 x + 20 tan x − 3 = 0 100 tan 2 x + 20 tan x − 3 = 0

2 cos 2 x − cos x + 15 = 0 2 cos 2 x − cos x + 15 = 0

20 sin 2 x − 27 sin x + 7 = 0 20 sin 2 x − 27 sin x + 7 = 0

2 tan 2 x + 7 tan x + 6 = 0 2 tan 2 x + 7 tan x + 6 = 0

130 tan 2 x + 69 tan x − 130 = 0 130 tan 2 x + 69 tan x − 130 = 0

For the following exercises, use a calculator to find all solutions to four decimal places.

sin x = 0.27 sin x = 0.27

sin x = −0.55 sin x = −0.55

tan x = −0.34 tan x = −0.34

cos x = 0.71 cos x = 0.71

For the following exercises, solve the equations algebraically, and then use a calculator to find the values on the interval [ 0 , 2 π ) . [ 0 , 2 π ) . Round to four decimal places.

tan 2 x + 3 tan x − 3 = 0 tan 2 x + 3 tan x − 3 = 0

6 tan 2 x + 13 tan x = −6 6 tan 2 x + 13 tan x = −6

tan 2 x − sec x = 1 tan 2 x − sec x = 1

sin 2 x − 2 cos 2 x = 0 sin 2 x − 2 cos 2 x = 0

2 tan 2 x + 9 tan x − 6 = 0 2 tan 2 x + 9 tan x − 6 = 0

4 sin 2 x + sin ( 2 x ) sec x − 3 = 0 4 sin 2 x + sin ( 2 x ) sec x − 3 = 0

For the following exercises, find all solutions exactly to the equations on the interval [ 0 , 2 π ) . [ 0 , 2 π ) .

csc 2 x − 3 csc x − 4 = 0 csc 2 x − 3 csc x − 4 = 0

sin 2 x − cos 2 x − 1 = 0 sin 2 x − cos 2 x − 1 = 0

sin 2 x ( 1 − sin 2 x ) + cos 2 x ( 1 − sin 2 x ) = 0 sin 2 x ( 1 − sin 2 x ) + cos 2 x ( 1 − sin 2 x ) = 0

3 sec 2 x + 2 + sin 2 x − tan 2 x + cos 2 x = 0 3 sec 2 x + 2 + sin 2 x − tan 2 x + cos 2 x = 0

sin 2 x − 1 + 2 cos ( 2 x ) − cos 2 x = 1 sin 2 x − 1 + 2 cos ( 2 x ) − cos 2 x = 1

tan 2 x − 1 − sec 3 x cos x = 0 tan 2 x − 1 − sec 3 x cos x = 0

sin ( 2 x ) sec 2 x = 0 sin ( 2 x ) sec 2 x = 0

sin ( 2 x ) 2 csc 2 x = 0 sin ( 2 x ) 2 csc 2 x = 0

2 cos 2 x − sin 2 x − cos x − 5 = 0 2 cos 2 x − sin 2 x − cos x − 5 = 0

1 sec 2 x + 2 + sin 2 x + 4 cos 2 x = 4 1 sec 2 x + 2 + sin 2 x + 4 cos 2 x = 4

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  • x^4-5x^2+4=0
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  • \log _2(x+1)=\log _3(27)
  • 3^x=9^{x+5}
  • What is the completing square method?
  • Completing the square method is a technique for find the solutions of a quadratic equation of the form ax^2 + bx + c = 0. This method involves completing the square of the quadratic expression to the form (x + d)^2 = e, where d and e are constants.
  • What is the golden rule for solving equations?
  • The golden rule for solving equations is to keep both sides of the equation balanced so that they are always equal.
  • How do you simplify equations?
  • To simplify equations, combine like terms, remove parethesis, use the order of operations.
  • How do you solve linear equations?
  • To solve a linear equation, get the variable on one side of the equation by using inverse operations.

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IMAGES

  1. Solving trig equations using a calculator (approximate solutions)

    solving trig equations with domain calculator

  2. solving trig equations on an interval

    solving trig equations with domain calculator

  3. Ex: Solve a Trigonometric Equation Using a Graphing Calculator

    solving trig equations with domain calculator

  4. Trigonometric Equation with Domain Change GCSE Level A

    solving trig equations with domain calculator

  5. Solving Trig Equations using a Graphics Calculator

    solving trig equations with domain calculator

  6. solving trig equations using a calculator

    solving trig equations with domain calculator

VIDEO

  1. Solving Trig Equations

  2. Solving trig equations (unit 2 revision sheet)

  3. Solving Trig Equations Using Identities

  4. Exact value of trig function (trig review for Calc 2/15)

  5. Solving Trig Equations

  6. Solving Harder Trig Equations

COMMENTS

  1. Trigonometric Equation Calculator

    What is tangent equal to? The tangent function (tan), is a trigonometric function that relates the ratio of the length of the side opposite a given angle in a right-angled triangle to the length of the side adjacent to that angle. How to solve trigonometric equations step-by-step?

  2. Trigonometric Equation Solver

    Mathematics Symbolic Computation Trigonometric Equation Solver Trig Equation Solving Trigonometric Equation to solve (one per line) 2cos (x)=pi/4 Variable (s) Solving Domain Set R (Reals) Z (Integers) C (Complex) Result Format Automatic Selection Exact Value (when possible) Approximate Numerical Value Scientific Notation Solve Equation

  3. Find the Domain Calculator

    Step 1: Enter the Function you want to domain into the editor. The domain calculator allows you to take a simple or complex function and find the domain in both interval and set notation instantly. Step 2: Click the blue arrow to submit and see the result!

  4. Trigonometric Equations Calculator & Solver

    Get detailed solutions to your math problems with our Trigonometric Equations step-by-step calculator. Practice your math skills and learn step by step with our math solver. Check out all of our online calculators here. 8sin ( x) = 2 + 4 csc ( x) Go! Math mode Text mode . ( ) / ÷ 2 √ √ ∞ e π ln log log lim d/dx D x ∫ ∫ | | θ = >

  5. 3.3: Solving Trigonometric Equations

    In other words, trigonometric equations may have an infinite number of solutions. Additionally, like rational equations, the domain of the function must be considered before we assume that any solution is valid. ... (\PageIndex{5B}\): Using a Calculator to Solve a Trigonometric Equation Involving Secant. Use a calculator to solve the equation ...

  6. Trigonometry Calculator

    Type a math problem Solve Related Concepts Trigonometry Trigonometry is a branch of mathematics concerned with relationships between angles and side lengths of triangles. In particular, the trigonometric functions relate the angles of a right triangle with ratios of its side lengths.

  7. Solving Trigonometric Equations

    Example 3: Solve the Trigonometric Equation in Linear Form. Solve the equation exactly: 2cosθ − 3 = − 5, 0 ≤ θ < 2π. Solve exactly the following linear equation on the interval [0, 2π): 2sinx + 1 = 0. Give your answers as exact values, as a list separated by commas.

  8. 9.6: Solving Trigonometric Equations

    Solve trigonometric equations using a calculator. Solve trigonometric equations that are quadratic in form. Solve trigonometric equations using fundamental identities. ... Additionally, like rational equations, the domain of the function must be considered before we assume that any solution is valid. The period of both the sine function and the ...

  9. Wolfram|Alpha Widgets: "Trigonometric Equations Solver"

    Get the free "Trigonometric Equations Solver" widget for your website, blog, Wordpress, Blogger, or iGoogle. Find more Education widgets in Wolfram|Alpha.

  10. Functions Domain Calculator

    To find the domain of a function, consider any restrictions on the input values that would make the function undefined, including dividing by zero, taking the square root of a negative number, or taking the logarithm of a negative number. Remove these values from the set of all possible input values to find the domain of the function.

  11. Trigonometric Equations Solver

    This calculator can solve basic trigonometric equations such as: or . The calculator will find exact or approximate solutions on custom range. Solution can be expressed either in radians or degrees. Trigonometric Equation Solver show help ↓↓ examples ↓↓ 0 1 2 3 4 5 6 7 8 9 - /. del ( · x ) = Type r to input square roots . Find:

  12. Mathway

    Free math problem solver answers your trigonometry homework questions with step-by-step explanations.

  13. Calculus I

    Let's just jump into the examples and see how to solve trig equations. Example 1 Solve 2cos(t) =√3 2 cos ( t) = 3 . Show Solution. Now, in a calculus class this is not a typical trig equation that we'll be asked to solve. A more typical example is the next one. Example 2 Solve 2cos(t) =√3 2 cos ( t) = 3 on [−2π,2π] [ − 2 π, 2 π] .

  14. Trig Calculator

    This is the Trig Calculator. Start by entering some numbers. Tip: You don't need to go from the top to the bottom. You can calculate anything, in any order. Trig Calculator Created by Luis Hoyos Reviewed by Davide Borchia Last updated: Jan 18, 2024 Cite Table of contents: The sine and cosine trigonometric functions

  15. Trigonometric equations online calculator

    f(trig(x)) = 0 where - some arbitrary function, trig(x) - some trigonometric function. As a rule, to solve trigonometric equation one need to transform it to the simplier form which has a known solution. The transformation can be done by using different trigonometric formulas . For example, consider the solution of trigonometric equation

  16. 3.6: Solving Trigonometric Equations

    In other words, trigonometric equations may have an infinite number of solutions. Additionally, like rational equations, the domain of the function must be considered before we assume that any solution is valid. ... (\PageIndex{5B}\): Using a Calculator to Solve a Trigonometric Equation Involving Secant. Use a calculator to solve the equation ...

  17. Trigonometric Equation Calculator

    This calculator will allow you to solve trig equations, showing all the steps of the way. All you need to do is to provide a valid trigonometric equation, with an unknown (x). It could be something simple as 'sin (x) = 1/2', or something more complex like 'sin^2 (x) = cos (x) + tan (x)'.

  18. Solving Trigonometric Equations

    Example 1: Solving Trig Equations in Radians. Let's start with a simple example and solve 2 s i n ( θ) = ( 2). The first step is to isolate the trig function: s i n ( θ) = ( 2) / 2.

  19. Trigonometric equations and identities

    In this unit, you'll explore the power and beauty of trigonometric equations and identities, which allow you to express and relate different aspects of triangles, circles, and waves. You'll learn how to use trigonometric functions, their inverses, and various identities to solve and check equations and inequalities, and to model and analyze problems involving periodic motion, sound, light, and ...

  20. 7.5 Solving Trigonometric Equations

    Solving Equations Involving a Single Trigonometric Function. When we are given equations that involve only one of the six trigonometric functions, their solutions involve using algebraic techniques and the unit circle (see Figure 2).We need to make several considerations when the equation involves trigonometric functions other than sine and cosine.

  21. Calculus I

    Section 1.4 : Solving Trig Equations. Without using a calculator find the solution (s) to the following equations. If an interval is given find only those solutions that are in the interval. If no interval is given find all solutions to the equation. 4sin(3t) = 2 4 sin. ⁡. ( 3 t) = 2 Solution. 4sin(3t) = 2 4 sin. ⁡.

  22. Equation Calculator

    Completing the square method is a technique for find the solutions of a quadratic equation of the form ax^2 + bx + c = 0. This method involves completing the square of the quadratic expression to the form (x + d)^2 = e, where d and e are constants. What is the golden rule for solving equations?