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Precalculus : Solving Trigonometric Equations and Inequalities

Study concepts, example questions & explanations for precalculus, all precalculus resources, example questions, example question #1 : solving trigonometric equations and inequalities.

how to solve trig equations and inequalities

Use the trigonometric identities to switch sec into terms of tan:

how to solve trig equations and inequalities

We begin by setting the right side of the equation equal to 0.

how to solve trig equations and inequalities

The equation might be easier to factor using the following substitution.

how to solve trig equations and inequalities

This gives the following

how to solve trig equations and inequalities

This can be factored as follows

how to solve trig equations and inequalities

Replacing our substitution therefore gives

how to solve trig equations and inequalities

Within our designated domain, we get three answers between our two equations.

how to solve trig equations and inequalities

Example Question #3 : Solving Trigonometric Equations And Inequalities

how to solve trig equations and inequalities

Begin by isolating the tangent side of the equation:

how to solve trig equations and inequalities

Next, take the inverse tangent of both sides:

how to solve trig equations and inequalities

Divide by five to get the final answer:

how to solve trig equations and inequalities

Example Question #4 : Solving Trigonometric Equations And Inequalities

Use trigonometric identities to solve for the angle value.

how to solve trig equations and inequalities

There are two ways to solve this problem. The first involves two trigonometric identities:

how to solve trig equations and inequalities

The second method allows us to only use the first trigonometric identity:

how to solve trig equations and inequalities

Example Question #5 : Solving Trigonometric Equations And Inequalities

Use trigonometric identities to solve the equation for the angle value.

how to solve trig equations and inequalities

Substituting this value into the original equation gives us:

how to solve trig equations and inequalities

Example Question #6 : Solving Trigonometric Equations And Inequalities

how to solve trig equations and inequalities

Some other identities that are important to know are:

how to solve trig equations and inequalities

Example Question #7 : Solving Trigonometric Equations And Inequalities

how to solve trig equations and inequalities

Set both terms equal to zero and solve.

how to solve trig equations and inequalities

Example Question #8 : Solving Trigonometric Equations And Inequalities

how to solve trig equations and inequalities

Example Question #9 : Solving Trigonometric Equations And Inequalities

how to solve trig equations and inequalities

Factor the left side of the equation.

how to solve trig equations and inequalities

The correct answer is:

how to solve trig equations and inequalities

There is no solution.

how to solve trig equations and inequalities

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Practice Quizzes

3.10 Trigonometric Equations and Inequalities

4 min read • january 29, 2023

Kashvi Panjolia

Kashvi Panjolia

Solving Trigonometric Inequalities

The inverse trigonometric functions , such as arcsine and arccosine , are used to undo the actions of regular trigonometric functions, such as sine and cosine. This means that if you have an equation or inequality that contains one of these regular trigonometric functions, you can use the corresponding inverse trigonometric function to solve for the variable in that equation or inequality.

For example, let's say we have the equation sin(x) = 0.5. To solve for x, we can use the inverse sine function, or arcsin, which would give us the solution x = 30 degrees. This is because the sine of 30 degrees is 0.5. In this way, we can use inverse trigonometric functions to find the value of a variable that would produce a specific result when plugged into a regular trigonometric function.

When it comes to solving a trigonometric inequality , the process is similar. For example, let's say we have the inequality sin(x) > 0.5. To solve this inequality, we can use the inverse sine function to find the values of x that would produce a result greater than 0.5 when plugged into the sine function. In this case, the solution would be x > 30 degrees and x < 150 degrees.

You can use the unit circle or your calculator to solve trigonometric inequalities and equations. Be sure to pay attention to whether you need to use degree mode or radian mode to find the correct answer.

Solving Trigonometric Equations

It's important to note that when using inverse trigonometric functions , the solutions you find may need to be modified due to domain restrictions . This means that there are certain values for which the inverse trigonometric functions are not defined, and you will have to take these restrictions into account when interpreting and using the solutions you find. For example, the function arcsin(x) only outputs values between -90 and 90 degrees. Therefore, if we get a solution that is outside this range, we would need to adjust it accordingly.

Due to the periodic nature of sine, cosine, and tangent , there are infinite solutions to these equations. For example, the equation above, sin(x) = 0.5, has a solution at 30 degrees, but it also has a solution at 390 degrees, 750 degrees, and so on because the sine function has a period of 2𝛑. This means that every 2𝛑 radians, or 360 degrees, the sine of the angle repeats. This is why ranges of the inverse trigonometric functions are restricted to only include one period of their respective trigonometric functions.

Let's solve the equation cos(x) = -0.2 by finding the value of x. To solve this equation, we can use the inverse cosine function, or arccosine , to find the value of x that would produce a result of -0.2 when plugged into the cosine function.

We first isolate the term cos(x) on one side of the equation: cos(x) = -0.2. In this case, it is already isolated. Next, we would take the arccosine of both sides: x = arccos(-0.2) to find the angle, x. Just like when you are solving the equation 3x = 6 by dividing both sides by 3 to find x, we are doing the same thing here because arccosine is the inverse of cosine, and division is the inverse of multiplication.

At this point, we would have a solution of x = arccos(-0.2). However, we must remember that the range of arccosine is between 0 and 180 degrees. Therefore, we need to restrict the domain of the solution in order to make sure that it falls within this valid range.

Image courtesy of Wiktionary.

In this case, the solution would be x = arccos(-0.2) = 163.4 degrees or x = 196.6 degrees. This means that the value of x that produces a cosine of -0.2 is 163.4 degrees and 196.6 degrees. We need to restrict our solution to x=163.4 degrees only because 196.6 degrees falls outside the range of the arccosine function. This example illustrates how it is important to be aware of the domain restrictions when working with inverse trigonometric functions and to adjust the solutions accordingly to ensure they fall within the valid range.

Domain and Range Restrictions on Trigonometric Functions

The same principles apply when solving equations and inequalities involving the sine and tangent functions. Below is a chart explaining the domain and range restrictions of the sine, cosine, and tangent functions, as well as their inverses. The square brackets indicate that the endpoints are included in the interval, and the open parentheses indicate that they are excluded.

Practice Problems

1. Which of the following is the correct solution to the equation sin(x) = 0.6, when restricted to the domain of 0 to 360 degrees?

A) 30 degrees

B) 150 degrees

C) 210 degrees

D) 330 degrees

Answer: B) 150 degrees

2. Which of the following is the correct solution to the inequality cos(x) > 0.5, when restricted to the domain of 0 to 360 degrees?

A) x < 30 degrees or x > 150 degrees

B) x > 30 degrees and x < 150 degrees

C) x < 30 degrees or x > 210 degrees

D) x > 30 degrees and x < 210 degrees

Answer: B) x > 30 degrees and x < 150 degrees

3. Which of the following is the correct solution to the equation tan(x) = -2, when restricted to the domain of -90 to 90 degrees?

A) -63.4 degrees

B) -116.6 degrees

C) 116.6 degrees

D) None of the above

Answer: B) -116.6 degrees

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How to Easily Solve Trigonometric Equations

Last Updated: February 1, 2023 Fact Checked

wikiHow is a “wiki,” similar to Wikipedia, which means that many of our articles are co-written by multiple authors. To create this article, 17 people, some anonymous, worked to edit and improve it over time. There are 10 references cited in this article, which can be found at the bottom of the page. This article has been fact-checked, ensuring the accuracy of any cited facts and confirming the authority of its sources. This article has been viewed 256,617 times. Learn more...

Did you get homework from your teacher that was about solving Trigonometric equations? Did you maybe not pay full attention in class during the lesson on Trigonometric questions? Do you even know what "Trigonometric" means? If you answered yes to these questions, then you don't need to worry because this wikiHow will teach you how to solve Trigonometric equations.

Step 1 Know the Solving concept.

  • To solve a trig equation, transform it into one or many basic trig equations. Solving trig equations finally results in solving 4 types of basic trig equations.

Step 2 Know how to solve basic trig equations.

  • There are 4 types of basic trig equations:
  • sin x = a ; cos x = a
  • tan x = a ; cot x = a
  • Solving basic trig equations proceeds by studying the various positions of the arc x on the trig circle, and by using trig conversion table (or calculator ). To fully know how to solve these basic trig equations, and similar, see book titled :"Trigonometry: Solving trig equations and inequalities" (Amazon E-book 2010).
  • Example 1. Solve sin x = 0.866. The conversion table (or calculator) gives the answer: x = Pi/3. The trig circle gives another arc (2Pi/3) that has the same sin value (0.866). The trig circle also gives an infinity of answers that are called extended answers.
  • x1 = Pi/3 + 2k.Pi, and x2 = 2Pi/3. (Answers within period (0, 2Pi))
  • x1 = Pi/3 + 2k Pi, and x2 = 2Pi/3 + 2k Pi. (Extended answers).
  • Example 2. Solve: cos x = -1/2. Calculators give x = 2 Pi/3. The trig circle gives another x = -2Pi/3.
  • x1 = 2Pi/3 + 2k.Pi, and x2 = - 2Pi/3. (Answers within period (0, 2Pi))
  • x1 = 2Pi/3 + 2k Pi, and x2 = -2Pi/3 + 2k.Pi. (Extended answers)
  • Example 3. Solve: tan (x - Pi/4) = 0.
  • x = Pi/4 ; (Answer)
  • x = Pi/4 + k Pi; ( Extended answer)
  • Example 4. Solve cot 2x = 1.732. Calculators and the trig circle give
  • x = Pi/12 ; (Answer)
  • x = Pi/12 + k Pi ; (Extended answers)

Step 3 Learn the Transformations used in solving trig equations.

  • To transform a given trig equation into basic trig ones, use common algebraic transformations ( factoring , common factor , polynomial identities...), definitions and properties of trig functions, and trig identities. There are about 31, among them the last 14 trig identities, from 19 to 31, are called Transformation Identities, since they are used in the transformation of trig equations. [4] X Research source See book mentioned above.
  • Example 5: The trig equation: sin x + sin 2x + sin 3x = 0 can be transformed, using trig identities, into a product of basic trig equations: 4cos x*sin (3x/2)*cos (x/2) = 0. The basic trig equations to be solved are: cos x = 0 ; sin (3x/2) = 0 ; and cos (x/2) = 0.

Step 4 Find the arcs whose trig functions are known.

  • Before learning solving trig equations, you must know how to quickly find the arcs whose trig functions are known. Conversion values of arcs (or angles) are given by trig tables or calculators. [6] X Research source
  • Example: After solving, get cos x = 0.732. Calculators give the solution arc x = 42.95 degree. The trig unit circle will give other solution arcs that have the same cos value.

Step 5 Graph the solution arcs on the trig unit circle.

  • You can graph to illustrate the solution arcs on the trig unit circle. The terminal points of these solution arcs constitute regular polygons on the trig circle. For examples:
  • The terminal points of the solution arcs x = Pi/3 + k.Pi/2 constitute a square on the trig unit circle.
  • The solution arcs x = Pi/4 + k.Pi/3 are represented by the vertexes of a regular hexagon on the trig unit circle.

Step 6 Learn the Approaches to solve trig equations.

  • A. Approach 1.
  • Transform the given trig equation into a product in the form: f(x).g(x) = 0 or f(x).g(x).h(x) = 0, in which f(x), g(x) and h(x) are basic trig equations.
  • Example 6. Solve: 2cos x + sin 2x = 0. (0 < x < 2Pi)
  • Solution. Replace in the equation sin 2x by using the identity: sin 2x = 2*sin x*cos x.
  • cos x + 2*sin x*cos x = 2cos x*( sin x + 1) = 0. Next, solve the 2 basic trig functions: cos x = 0, and (sin x + 1) = 0.
  • Example 7. Solve: cos x + cos 2x + cos 3x = 0. (0 < x < 2Pi)
  • Solution: Transform it to a product, using trig identities: cos 2x(2cos x + 1 ) = 0. Next, solve the 2 basic trig equations: cos 2x = 0, and (2cos x + 1) = 0.
  • Example 8. Solve: sin x - sin 3x = cos 2x. (0 < x < 2Pi)
  • B. Approach 2.
  • Transform the given trig equation into a trig equation having only one unique trig function as variable. There are a few tips on how to select the appropriate variable. The common variables to select are: sin x = t; cos x = t; cos 2x = t, tan x = t and tan (x/2) = t.
  • Example 9. Solve: 3sin^2 x - 2cos^2 x = 4sin x + 7 (0 < x < 2Pi).
  • Solution. Replace in the equation (cos^2 x) by (1 - sin^2 x), then simplify the equation:
  • 3sin^2 x - 2 + 2sin^2 x - 4sin x - 7 = 0. Call sin x = t. The equation becomes: 5t^2 - 4t - 9 = 0. This is a quadratic equation that has 2 real roots: t1 = -1 and t2 = 9/5. The second t2 is rejected since > 1. Next, solve: t = sin = -1 --> x = 3Pi/2.
  • Example 10. Solve: tan x + 2 tan^2 x = cot x + 2.
  • Solution. Call tan x = t. Transform the given equation into an equation with t as variable: (2t + 1)(t^2 - 1) = 0. Solve for t from this product, then solve the basic trig equation tan x = t for x.

Step 7 Solve special types of trig equations.

  • There are a few special types of trig equations that require some specific transformations. Examples:
  • a*sin x+ b*cos x = c ; a(sin x + cos x) + b*cos x*sin x = c ;
  • a*sin^2 x + b*sin x*cos x + c*cos^2 x = 0

Step 8 Learn the Periodic Property of trig functions.

  • The function f(x) = sin x has 2Pi as period.
  • The function f(x) = tan x has Pi as period.
  • The function f(x) = sin 2x has Pi as period.
  • The function f(x) = cos (x/2) has 4Pi as period.
  • If the period is specified in the problem/test, you have to only find the solution arc(s) x within this period.
  • NOTE: Solving trig equation is a tricky work that often leads to errors and mistakes. Therefore, answers should be carefully checked. After solving, you can check the answers by using a graphing calculator to directly graph the given trig equation R(x) = 0. The answers (real roots) will be given in decimals. For example, Pi is given by the value 3.14
  • ↑ https://www.khanacademy.org/math/geometry/hs-geo-trig/hs-geo-solve-for-a-side/a/unknown-side-in-right-triangle-w-trig
  • ↑ https://www.purplemath.com/modules/solvtrig.htm
  • ↑ https://www.shelovesmath.com/algebra/advanced-algebra/parent-graphs-and-transformations/#GenericTransformationsofFunctions
  • ↑ https://www.khanacademy.org/math/precalculus/trig-equations-and-identities-precalc/using-trig-identities-precalc/v/examples-using-pythagorean-identities-to-simplify-trigonometric-expressions
  • ↑ https://www.khanacademy.org/math/precalculus/trig-equations-and-identities-precalc/inverse-trig-functions-precalc/v/inverse-trig-functions-arcsin
  • ↑ https://www.mathopenref.com/arcsin.html
  • ↑ https://courses.lumenlearning.com/precalculus/chapter/unit-circle-sine-and-cosine-functions/
  • ↑ https://mathbitsnotebook.com/Algebra2/TrigConcepts/TCEquationsMore.html
  • ↑ https://www.analyzemath.com/trigonometry/properties.html
  • ↑ https://www.youtube.com/watch?v=8Z60_yXX4xA

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  • \sin (x)+\sin (\frac{x}{2})=0,\:0\le \:x\le \:2\pi
  • \cos (x)-\sin (x)=0
  • \sin (4\theta)-\frac{\sqrt{3}}{2}=0,\:\forall 0\le\theta<2\pi
  • 2\sin ^2(x)+3=7\sin (x),\:x\in[0,\:2\pi ]
  • 3\tan ^3(A)-\tan (A)=0,\:A\in \:[0,\:360]
  • 2\cos ^2(x)-\sqrt{3}\cos (x)=0,\:0^{\circ \:}\lt x\lt 360^{\circ \:}
  • 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?
  • To solve a trigonometric simplify the equation using trigonometric identities. Then, write the equation in a standard form, and isolate the variable using algebraic manipulation to solve for the variable. Use inverse trigonometric functions to find the solutions, and check for extraneous solutions.
  • What is a basic trigonometric equation?
  • A basic trigonometric equation has the form sin(x)=a, cos(x)=a, tan(x)=a, cot(x)=a
  • How to convert radians to degrees?
  • The formula to convert radians to degrees: degrees = radians * 180 / π
  • What is cotangent equal to?
  • The cotangent function (cot(x)), is the reciprocal of the tangent function.cot(x) = cos(x) / sin(x)

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  • High School Math Solutions – Trigonometry Calculator, Trig Equations In a previous post, we learned about trig evaluation. It is important that topic is mastered before continuing... Read More

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10.7: Trigonometric Equations and Inequalities

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  • Page ID 119200

Solving "Basic" Trigonometric Equations

In Sections 10.2 , 10.3 , and most recently 10.6 , we solved some basic equations involving trigonometric functions. In this section, we expand our skill set to handle a broader variety of equations involving trigonometric functions. The base tactic employs two critical elements from Trigonometry - reference angles and quadrants. 1

Strategy for Solving "Basic" Trigonometric Equations

Given any trigonometric equation of the form \[ \text{trig}(u) = c, \label{trigeqn}\]where "trig" could be \( \sin, \cos, \tan, \csc, \sec, \) or \( \cot \), do the following:

  • If the value of \( c \) is such that \( u \) is an axial angle, skip to Step 5.
  • Find the reference angle, \( \hat{u} \).
  • Use the sign of \( c \) to determine in which quadrants \( u \) terminates.
  • Using the reference angle from Step 2, write the solutions associated with the quadrants listed in Step 3. If possible, compress the solution set to a single equation (this is most often the case with equations involving the tangent or the cotangent).
  • If \( u \) is a function of another variable, solve the equations in Step 4 (or Step 1, if the angle is an axial angle) for the "true" variable.
  • Check for extraneous solutions!

A few notes about this strategy are in order and I will focus on showcasing how to solve \( \cos(3x) = \frac{\sqrt{3}}{2} \) while making my remarks.

First, I am using the variable \( u \) for the argument of the trigonometric function instead of the traditional "angular" variables \( t \), \( \alpha \), \( \beta \), \( \theta \), or \( x \). This is purposeful because, given a more complex trigonometric equation, we will often need to perform a \( u \)-substitution to get the equation into a more "basic" form. For example, given \( \cos(3x) = \frac{\sqrt{3}}{2} \), we will perform a \( u \)-substitution by letting \( u = 3x \). This simplifies our equation to the more "basic" form, \[\cos\left(u \right) = \frac{\sqrt{3}}{2}.\nonumber\]

To determine the reference angle, \( \hat{u} \), needed so that \( \cos(\hat{u}) = \frac{\sqrt{3}}{2} \), we refer to our special triangles. In this case, we find \( \hat{u} = \frac{\pi}{6} \).

Second, since all trigonometric functions are positive in QI and one other quadrant, there will be two quadrants in which the angle \( u \) will terminate if \( c \gt 0 \). A similar argument shows that there will be two quadrants in which \( u \) terminates if \( c \lt 0 \). In our current example, \( \cos(u) \) is positive. Therefore, \( u \) will terminate in QI and QIV (the quadrants where cosine is positive).

Using the reference angle, \( \hat{u} = \frac{\pi}{6} \), we write the solutions associated with quadrants I and IV.\[ \begin{array}{lrclclcl} \text{Solutions associated with QI: } & u & = & \hat{u} + 2\pi k & = & \frac{\pi}{6} + 2 \pi k & & \\ \text{Solutions associated with QIV: } & u & = & 2\pi - \hat{u} + 2\pi k & = & 2\pi - \frac{\pi}{6} + 2 \pi k & = & \frac{11\pi}{6} + 2 \pi k, \\ \end{array} \nonumber \]

where \( k \in \mathbb{Z} \). The addition of \( 2 \pi k \) represents the fact that any full rotation of \( 2\pi \) lands the terminating side of \( u \) at the same spot. I leave it to the reader to verify the given solutions do indeed terminate in QI and QIV.

Our task in the original equation was to solve for \( x \) - not \( u \). Substituting \( u = 3x \) into our solutions, we get\[ \begin{array}{lrcl} \text{Solutions associated with QI: } & 3x & = & \frac{\pi}{6} + 2 \pi k & & \\ \text{Solutions associated with QIV: } & 3x & = & \frac{11\pi}{6} + 2 \pi k. \\ \end{array} \nonumber \]

Finally, dividing both sides of each equation by \( 3 \), 2 we arrive at our solutions.\[ \begin{array}{lrcl} \text{Solutions associated with QI: } & x & = & \frac{\pi}{18} + \frac{2 \pi}{3} k & & \\ \text{Solutions associated with QIV: } & x & = & \frac{11\pi}{18} + \frac{2 \pi}{3} k. \\ \end{array} \nonumber \]

Notice that \( \frac{11 \pi}{18} \) does not terminate in QIV yet it is supposed to be associated with QIV. Did we do something wrong? This leads to my third note. In Step 4, I use the word " associated " because, as was mentioned previously, \( u \) is generally a substitution for a more complex variable expression. In our current example, we were originally given \( \cos(3x) = \frac{\sqrt{3}}{2} \). We let \( u = 3x \) and solved \( \cos(u) = \frac{\sqrt{3}}{2} \) to get \( u = \frac{\pi}{6} + 2 \pi k \) and \( u = \frac{11 \pi}{6} + 2 \pi k \). These angles occur in QI and QIV, respectively (where the cosine is positive). So far, we should feel good about our results; however , QI and QIV are the quadrants where \( u \) terminates - not \( x \). Since we are supposed to be solving for \( x \), it is often the case that the angular frequency (the coefficient on \( x \)) will change the final quadrants of the solutions. This is absolutely fine!

For example, while \( x = \frac{11 \pi}{18} \in \text{QII} \neq \text{QIV} \), substituting this into our original equation, we get\[ \cos{\left( 3 \left( \frac{11 \pi}{18} \right) \right)} = \cos{\left( \frac{11 \pi}{6} \right)}. \nonumber \]Our final argument, \( \frac{11 \pi}{6} \), is actually an angle in QIV.

We will repeat this technique in the example below.

Example \( \PageIndex{1} \)

Solve the following equations and check your answers analytically. List the solutions which lie in the interval \([0,2\pi)\) and verify them analytically (i.e., by hand) or by using graphing technology.

  • \(\sin(2x) = -\frac{\sqrt{3}}{2}\)
  • \(\csc\left(\frac{1}{3}x-\pi \right) = \sqrt{2}\)
  • \(\cot\left(3x \right) = 0\)
  • \(\sec^{2}(x) = 4\)
  • \(\tan\left(\frac{x}{2}\right) = -3\)
  • \(\sin(2x) = 0.87\)

Precalc_10_7_Fig1a.png

  • This looks like it might be more challenging than part 1, but that's an illusion. We begin by simplifying matters and let \( u = \frac{1}{3}x - \pi \). Our equation now has the form \[\csc(u) = \sqrt{2}.\nonumber\]\( \sqrt{2} = \frac{\sqrt{2}}{1} \) is a special ratio, and referring to our special triangles, we see that the cosecant of \( \frac{\pi}{4} \) is \( \sqrt{2} \). Therefore, our reference angle is \( \hat{u} = \frac{\pi}{4} \). Moreover, since \( \csc(u) \) is positive, we are going to limit our initial work to quadrants I and II (where the cosecant is positive). \[ \begin{array}{lrclclcl} \text{Solutions associated with QI: } & u & = & \hat{u} + 2\pi k & = & \frac{\pi}{4} + 2 \pi k & & \\ \text{Solutions associated with QII: } & u & = & \pi - \hat{u} + 2\pi k & = & \pi - \frac{\pi}{4} + 2 \pi k & = & \frac{3 \pi}{4} + 2 \pi k, \\ \end{array} \nonumber \]for \( k \in \mathbb{Z} \). Resubsituting in for \( u \), we get\[ \begin{array}{lrcl} \text{Solutions associated with QI: } & \frac{1}{3}x - \pi & = & \frac{\pi}{4} + 2 \pi k \\ \text{Solutions associated with QII: } & \frac{1}{3}x - \pi & = & \frac{3 \pi}{4} + 2 \pi k. \\ \end{array} \nonumber \]These are both just basic algebra problems. Adding \( \pi \) to both sides, we get\[ \begin{array}{lrcl} \text{Solutions associated with QI: } & \frac{1}{3}x & = & \frac{5 \pi}{4} + 2 \pi k \\ \text{Solutions associated with QII: } & \frac{1}{3}x & = & \frac{7 \pi}{4} + 2 \pi k. \\ \end{array} \nonumber \]A common error is to treat the "\(2\pi k\)" and "\(\pi\)" terms as "like" terms and try to combine them when they are not. 3  Finally, we multiply both sides of each equation by \( 3 \), we get \[ \begin{array}{lrcl} \text{Solutions associated with QI: } & x & = & \frac{15 \pi}{4} + 6 \pi k \\ \text{Solutions associated with QII: } & x & = & \frac{21 \pi}{4} + 6 \pi k. \\ \end{array} \nonumber \]Let's check our work analytically. To check the first family of answers, we substitute, combine like terms, and simplify.\[\begin{array}{rclr} \csc\left(\frac{1}{3} \left[ \frac{15\pi}{4} + 6 \pi k \right] - \pi \right) & = & \csc\left(\frac{5\pi}{4} + 2\pi k - \pi \right) & \\ & = & \csc\left(\frac{\pi}{4} + 2\pi k\right) & \\ & = & \csc\left(\frac{\pi}{4}\right) & \left(\text{the period of cosecant is }2\pi \right) \\ & = & \sqrt{2} & \\ \end{array}\nonumber\]The family \(x= \frac{21\pi}{4} + 6 \pi k\) checks similarly. Despite having infinitely many solutions, we find that none of them lie in \([0,2\pi)\).

Precalc_10_7_Fig1c.png

  • The equation \(\tan\left(\frac{x}{2}\right) = -3\) has the form \(\tan(u) = -3\), where \( u = \frac{x}{2} \). We have reached a case where the ratio in question is not from a special triangle nor does it represent the result of \( u \) being an axial angle. How do we proceed? If \( \hat{u} \) is the reference angle, then \( \tan(\hat{u}) = |-3| = 3 \). 5  Therefore, \[\hat{u} = \arctan(3).\nonumber\]Since \( \tan(u) \) is negative, we will restrict our search for solutions to quadrants II and IV (where the tangent is negative). \[ \begin{array}{lrclcl} \text{Solutions associated with QII: } & u & = & \pi - \hat{u} + 2\pi k & = & \pi - \arctan(3) + \pi k \\ \text{Solutions associated with QIV: } & u & = & 2 \pi - \hat{u} + 2\pi k & = & 2 \pi - \arctan(3) + \pi k, \\ \end{array} \nonumber \]for \( k \in \mathbb{Z} \). Notice that we added \( \pi k \) instead of \( 2 \pi k \). This is because the tangent function (and the cotangent function, for that matter) repeat their behavior every \( \pi \) radians. Looking closely at these solutions, we can see that writing both is a bit overkill because when \( k = 1 \), the first solution becomes identical to the second solution at \( k = 0 \). Hence, we compress the equations into\[ \begin{array}{lrclcl} \text{Solutions associated with QII and QIV: } & u & = & \pi - \arctan(3) + \pi k. \\ \end{array} \nonumber \]An oddity occurs here that represents our need to not fall asleep at the proverbial mathematical wheel. The terms \( \pi \) and \(  \pi k\) can be grouped to become \( \pi + \pi k = \pi (1 + k) \). Since \( k \) is any integer, we can replace \( 1 + k \) with \( j \), where \( j \in \mathbb{Z} \). Hence, \[ \begin{array}{lrclcl} \text{Solutions associated with QII and QIV: } & u & = & - \arctan(3) + \pi j, \\ \end{array} \nonumber \]where \( j \in \mathbb{Z} \). Resubstituting \( \frac{x}{2} = u \), we get\[ \begin{array}{lrclcl} \text{Solutions associated with QII and QIV: } & \frac{x}{2} & = & - \arctan(3) + \pi j. \\ \end{array} \nonumber \]Multiplying both sides of this equation by \( 2 \), we finally arrive at\[ \begin{array}{lrclcl} \text{Solutions associated with QII and QIV: } & x & = & - 2\arctan(3) + 2\pi j. \\ \end{array} \nonumber \]To check our solutions, we note \[\begin{array}{rclr} \tan\left(\frac{-2\arctan(3) + 2\pi j}{2}\right) & = & \tan\left( -\arctan(3) + \pi j \right) & \\ & = & \tan\left(-\arctan(3) \right) & (\text{the period of tangent is }\pi) \\ & = & -\tan\left(\arctan(3) \right) & (\text{Even/Odd Identities}) \\ & = & -3 & (\text{See Theorem 10.6.2}) \\ \end{array}\nonumber\]To determine which of our answers lie in the interval \([0,2\pi)\), we first need to get an idea of the value of \(-2\arctan(3)\). While we could easily find an approximation using a calculator, 6 we proceed analytically. Since \(3 \gt 0\), it follows that \[0 \lt \arctan(3) \lt \frac{\pi}{2}.\nonumber\]Multiplying through by \(-2\) gives \[0 \gt -2\arctan(3) \gt -\pi.\nonumber\]We are now in a position to argue which of the solutions \(x = -2\arctan(3) + 2\pi j\) lie in \([0,2\pi)\). For \(j = 0\), we get \(x = -2\arctan(3) \lt 0\), so we discard this answer and all answers \(x = -2\arctan(3) + 2\pi j\) where \(j \lt 0\). Next, we turn our attention to \(j = 1\) and get \(x = -2\arctan(3) + 2\pi\). Starting with the inequality \(-\pi \lt -2\arctan(3) \lt 0\), we add \(2\pi\) and get \(\pi < -2\arctan(3) +2\pi < 2\pi\). This means \(x = -2\arctan(3) + 2\pi\) lies in \([0,2\pi)\). Advancing \(j\) to \(2\) produces \(x = -2\arctan(3) + 4\pi\). Once again, we get from \(-\pi < -2\arctan(3) < 0\) that \(3\pi < -2\arctan(3) + 4\pi < 4\pi\). Since this is outside the interval \([0,2\pi)\), we discard \(x = -2\arctan(3) + 4\pi\) and all solutions of the form \(x = -2\arctan(3) + 2\pi j\) for \(j > 2\). Graphically, we see \(y = \tan\left(\frac{x}{2}\right)\) and \(y = -3\) intersect only once on \([0,2\pi)\) at \(x = -2\arctan(3) + 2\pi\approx 3.7851\).

Screen Shot 2022-05-17 at 12.35.50 AM.png

Subsection Footnotes

1 As has already been mentioned, understanding how impactful reference angles and quadrants are in Trigonometry will allow you to easily master the subject.

2  Don’t forget to divide \(2 \pi k\) by 3 as well!

3 Do you see why?

4  The reader is encouraged to see what happens if we had chosen the reciprocal identity \(\cot (3 x)=\frac{1}{\tan (3 x)}\) instead. The graph on the calculator appears identical, but what happens when you try to find the intersection points?

5 Reference angles are always between \( 0 \) and \( \frac{\pi}{2} \) radians (inclusive). Therefore, any trigonometric function evaluated at a reference angle will be positive!

6  Your instructor will let you know if you should abandon the analytic route at this point and use your calculator. But seriously, what fun would that be?

Solving Equations Involving Different Trigonometric Functions Having the Same Angular Frequencies

Each of the problems in Example \( \PageIndex{1} \) was ideal in that it featured one trigonometric function. If, for example, we were faced with solving \[\sec^{2}(x) = \tan(x) + 3,\nonumber\]what would we do? If our ultimate goal is to refine a given trigonometric equation to the form of Equation \( \ref{trigeqn} \), how can we do this if the equation involves different trigonometric functions?

The answer to these questions comes from a mixture of trigonometric identities and algebraic manipulations. In this section, we see (via example) that, given an equation involving two different trigonometric functions with the same angular frequency , 7 we will need to use trigonometric identities and Algebra to simplify the equation to a form involving either a single trigonometric function or the product of different trigonometric functions equaling zero.

Example \( \PageIndex{2} \)

Solve the following equations and list the solutions which lie in the interval \([0,2\pi)\) and verify them analytically (i.e., by hand) or by using graphing technology.

  • \(3\cos(x) \sin^{2}(x) = \sin^{2}(x)\)
  • \(\sec^{2}(x) = \tan(x) + 3\)
  • We resist the temptation to divide both sides of \(3\cos(x)\sin^{2}(x) = \sin^{2}(x)\) by \(\sin^{2}(x)\) (recall, we have stated repeatedly that dividing by a variable expression implies you are assuming the expression is nonzero, which might not be the case) and instead gather all of the terms to one side of the equation and factor.\[\begin{array}{rclr} 3\cos(x) \sin^{2}(x) & = & \sin^{2}(x) & \\ 3\cos(x) \sin^{2}(x) - \sin^{2}(x) & = & 0 & \\ \sin^{2}(x) (3 \cos(x) - 1) & = & 0 & \left( \text{Factor out }\sin^{2}(x)\text{ from both terms}\right) \\ \end{array}\nonumber\]We get \[\sin^{2}(x) = 0 \qquad \text{or} \qquad 3\cos(x) - 1 = 0.\nonumber\]Solving each, we find \[\sin(x) = 0 \qquad \text{or} \qquad \cos(x) = \frac{1}{3}.\nonumber\]These equations are both of the form of Equation \( \ref{trigeqn} \) so solving them will not be too bad. The right side of \( \sin(x) = 0 \) implies an axial angle. The solution to this equation is \(x = \pi k\), with \(x = 0\) and \(x = \pi\) being the two solutions which lie in \([0,2\pi)\). To solve \(\cos(x) = \frac{1}{3}\), we note that the reference angle is \(\hat{x} = \arccos\left(\frac{1}{3}\right)\). Since the cosine is positive, we know that the quadrants in question are QI and QIV. Hence, the two families of solutions are \[ \begin{array}{lrclclcl} \text{Solutions associated with QI: } & x & = & \hat{x} + 2\pi k & = & \arccos\left(\frac{1}{3}\right) + 2 \pi k & & \\ \text{Solutions associated with QIV: } & x & = & 2\pi - \hat{x} + 2\pi k & = & 2\pi - \arccos\left(\frac{1}{3}\right) + 2 \pi k & = & -\arccos\left(\frac{1}{3}\right) + 2 \pi j, \\ \end{array} \nonumber \]where \( j, k \in \mathbb{Z} \). Note that I have, yet again, rewritten \( 2 \pi + 2 \pi k = 2 \pi ( 1 + k) = 2 \pi j\). We find the two solutions here which lie in \([0,2\pi)\) to be \(x = \arccos\left(\frac{1}{3}\right)\) and \(x = 2\pi - \arccos\left(\frac{1}{3}\right)\). We leave it to the reader to check these solutions graphically. Thus, the solutions to the original equation on the interval \( [0, 2 \pi) \) are \[ x = 0, \arccos\left(\frac{1}{3}\right), \pi, \text{ and } 2\pi - \arccos\left(\frac{1}{3}\right). \nonumber \]
  • The \( \sec^2(x) \) in an equation with \( \tan(x) \) screams "Pythagorean Identity!" \[\begin{array}{rclr} \sec^{2}(x) & = & \tan(x) + 3 & \\ 1 + \tan^{2}(x) & = & \tan(x) + 3 & \left( \text{Pythagorean Identity} \right) \\ \tan^{2}(x) - \tan(x) -2 & = & 0 & \\ u^2 - u - 2 & = & 0 & \left( u-\text{substitution} \right) \\ (u + 1)(u - 2) & = & 0 & \\ \end{array}\nonumber\]This gives \(u = -1\) or \(u = 2\). Since \(u = \tan(x)\), we have \(\tan(x) = -1\) or \(\tan(x) = 2\). Considering \(\tan(x) = -1\), we recognize the reference angle to be \( \frac{\pi}{4} \) and the quadrants to be QII and QIV.\[ \begin{array}{lrclclcl} \text{Solutions associated with QII and QIV: } & x & = & \pi - \hat{x} + \pi k & = & \pi - \frac{\pi}{4} + \pi k & = & - \frac{\pi}{4} + \pi j \\ \end{array} \nonumber \]where \( j = k + 1 \) and \( k \in \mathbb{Z} \). Instead of checking these solutions immediately, let's restrict our potential  solutions to the interval \( [0 , 2 \pi) \). 8  For \( j = 0 \), we get \( x = -\frac{\pi}{4} \), which will not work. Moving on to \( j = 1 \) and \( j = 2 \), we get \( x = \frac{3 \pi}{4} \) and \( x = \frac{7 \pi}{4} \), respectively. Any further, and our values for \( x \) will go beyond the requested interval.  Checking these refined solutions, we get\[ \begin{array}{lrcl} \text{For }x = \frac{3 \pi}{4} & & & \\ & \sec^2{\left( \frac{3 \pi}{4} \right)} & \overset{\text{?}}{=} & \tan{ \left( \frac{3 \pi}{4} \right) } + 3 \\ \implies & \left( -\sqrt{2} \right)^2 & \overset{\text{?}}{=} & -1 + 3 \\ \implies & 2 & \overset{\text{?}}{=} & 2 \\ \end{array} \nonumber \]Thus, we can see \( x = \frac{3 \pi}{4} \) works.\[ \begin{array}{lrcl} \text{For }x = \frac{7 \pi}{4} & & & \\ & \sec^2{\left( \frac{7 \pi}{4} \right)} & \overset{\text{?}}{=} & \tan{ \left( \frac{7 \pi}{4} \right) } + 3 \\ \implies & \left( \sqrt{2} \right)^2 & \overset{\text{?}}{=} & -1 + 3 \\ \implies & 2 & \overset{\text{?}}{=} & 2 \\ \end{array} \nonumber \]We see that \( x = \frac{7 \pi}{4} \) works as well. Now we look at \(\tan(x) = 2\). We should immediately recognize the right-hand side is  not  a special ratio for the tangent function. Therefore, we set the reference angle to \( \hat{x} = \arctan(|2|) = \arctan(2) \). Since the tangent function is positive in this case, the quadrants of interest are QI and QIII.\[ \begin{array}{lrclcl} \text{Solutions associated with QI and QIII: } & x & = & \hat{x} + \pi k & = & \arctan(2) + \pi k \\ \end{array} \nonumber \]where \( k \in \mathbb{Z} \). On \( \left[ 0, 2\pi \right) \), \( x = \arctan(2) \) and \( x = \arctan(2) + \pi \) are candidate solutions. Checking, we get\[ \begin{array}{lrcl} \text{For }x = \arctan(2) & & & \\ & \sec^2{\left( \arctan(2) \right)} & \overset{\text{?}}{=} & \tan{ \left( \arctan(2) \right) } + 3 \\ \end{array} \nonumber \]We need to reach back to Section 10.6 to help us with both the left- and right-hand sides of this equation. The right-hand side quickly simplifies to \( 2 + 3 = 5 \); however, the left-hand side requires us to let \( \hat{x} = \arctan(2) \implies \tan(\hat{x}) = 2 \). Luckily, \[ \sec^2{ \left( \arctan(2) \right) } = \sec^2{(\hat{x})} =  1 + \tan^2{(\hat{x})} = 1 + 2^2 = 5 \nonumber \]Therefore, the two sides  are , in fact, equivalent.\[ \begin{array}{lrclr} \text{For }x = \arctan(2) + \pi & & & & \\ & \sec^2{\left( \arctan(2) + \pi \right)} & \overset{\text{?}}{=} & \tan{ \left( \arctan(2) + \pi \right) } + 3 & \\ \implies & 1 + \tan^2{\left( \arctan(2) + \pi \right)} & \overset{\text{?}}{=} & \tan{ \left( \arctan(2) \right) } + 3 & \left( \text{the period of tangent is }\pi \text{ and Pythagorean Identity} \right) \\ \implies & 1 + \tan^2{\left( \arctan(2) \right) } & \overset{\text{?}}{=} & 5 & \left( \text{Reciprocal Identity and the period of tangent is }\pi \right) \\ \implies & 1 + 2^2 & \overset{\text{?}}{=} & 5 &  \\ \end{array} \nonumber \]Thus, \( x = \arctan(2) + \pi \) works as well. Hence, the solution set to the original equation on the interval \( \left[ 0, 2 \pi \right) \) is \[ x = \arctan(2), \frac{3 \pi}{4}, \arctan(2) + \pi, \text{ and } \frac{7 \pi}{4}. \nonumber \]

7 Recall from Section 10.5 that the angular frequency of a sinusoidal function is the coefficient of \( x \). That is, the angular frequency of \( f(x) = -2 \sin{(3x + 1)} - 4 \) is \( 3 \). In this section, we are slightly abusing the language "angular frequency" by allowing ourselves to include the non-sinusoidal trigonometric functions in this discussion. 8  I find it much better to get my "families of solutions" first, restrict them to the requested interval,  then  check solutions. It's a personal preference that, at times, can be inefficient; however, it avoids mistakes in the long run.

Solving Equations Involving Trigonometric Functions Having Different Angular Frequencies

If an equation contains more than one trigonometric function (whether the same or not) but with different angular frequencies, we will need to use the Double-Angle Identities (or, in rare cases, the Sum-to-Product or Product-to-Sum Formulas) and Algebra to simplify the equation to simpler equations of the form of Equation \( \ref{trigeqn} \).

Example \(\PageIndex{3}\)

Solve the following equations and list the solutions which lie in the interval \([0,2\pi)\). Verify your solutions on \([0,2\pi)\) graphically.

  • \(\cos(2x) = 3\cos(x) - 2\)
  • \(\cos(3x) = 2- \cos(x)\)
  • \(\cos(3x) = \cos(5x)\)
  • \(\sin(2x) =\sqrt{3} \cos(x)\)
  • \(\sin(x)\cos\left(\frac{x}{2}\right) + \cos(x)\sin\left(\frac{x}{2}\right) = 1\)
  • In the equation \(\cos(2x) = 3\cos(x) - 2\), we have the same circular function, namely cosine, on both sides but the arguments differ. Using the Double-Angle Identity \[\cos(2x) = 2\cos^{2}(x) - 1,\nonumber\]we obtain a "quadratic in disguise" and proceed as we have done in the past.\[\begin{array}{rclr} \cos(2x) & = & 3\cos(x) - 2 & \\ 2\cos^{2}(x) -1 & = & 3\cos(x) -2 & \left( \text{Double-Angle Identity} \right) \\ 2\cos^{2}(x) - 3\cos(x) +1 & = & 0 & \\ 2 u^2 - 3 u + 1 & = & 0 & \left( u-\text{Substitution} \right) \\ (2u - 1)(u - 1) & = & 0 & \\ \end{array}\nonumber\]This gives \(u = \frac{1}{2}\) or \(u = 1\). Since \(u = \cos(x)\), we get \(\cos(x) = \frac{1}{2}\) or \(\cos(x) = 1\). Solving \(\cos(x) = \frac{1}{2}\), we get \(\hat{x} = \frac{\pi}{3}\), and \( x \) is in QI or QIV.\[ \begin{array}{lrclclcl} \text{Solutions associated with QI: } & x & = & \hat{x} + 2\pi k & = & \frac{\pi}{3} + 2 \pi k & & \\ \text{Solutions associated with QIV: } & x & = & 2\pi - \hat{x} + 2\pi k & = & 2\pi - \frac{\pi}{3} + 2 \pi k & = & \frac{5 \pi}{3} + 2 \pi k, \\ \end{array} \nonumber \]where \( k \in \mathbb{Z} \). Solving \( \cos(x) = 1 \), we get \( x = 0 + 2 \pi k = 2 \pi k \). The answers which lie in \([0,2\pi)\) are \(x =0\), \(\frac{\pi}{3}\), and \(\frac{5\pi}{3}\). Graphing \(y = \cos(2x)\) and \(y = 3\cos(x) - 2\), we find, after a little extra effort, that the curves intersect in three places on \([0,2\pi)\), and the \(x\)-coordinates of these points confirm our results.

Screen Shot 2022-05-17 at 1.02.48 AM.png

  • While we could approach \(\cos(3x) = \cos(5x)\) in the same manner as we did the previous problem, we choose instead to showcase the utility of the Sum-to-Product Formulas. From \(\cos(3x) = \cos(5x)\), we get \(\cos(5x) - \cos(3x) = 0\), and it is the presence of \(0\) on the right hand side that indicates a switch to a product would be a good move. 9  Using the Sum-to-Product Formulas , we have that \[\cos(5x) - \cos(3x) = - 2 \sin\left( \frac{5x + 3x}{2}\right)\sin\left( \frac{5x - 3x}{2}\right) = -2 \sin(4x)\sin(x).\nonumber\]Hence, the equation \(\cos(5x) = \cos(3x)\) is equivalent to \(-2 \sin(4x) \sin(x) = 0\). From this, we get \(\sin(4x) = 0\) or \(\sin(x)\) = 0. Solving \(\sin(4x) = 0\) gives \(x = \frac{\pi}{4} k\) for integers \(k\) (notice how we are starting to speed up?), and the solution to \(\sin(x) = 0\) is \(x = \pi k\) for integers \(k\). The second set of solutions is contained in the first set of solutions, 10  so our final solution to \(\cos(5x) = \cos(3x)\) is \(x = \frac{\pi}{4} k\) for integers \(k\). There are eight of these answers which lie in \([0,2\pi)\): \(x = 0\), \(\frac{\pi}{4}\), \(\frac{\pi}{2}\), \(\frac{3\pi}{4}\), \(\pi\), \(\frac{5\pi}{4}\), \(\frac{3\pi}{2}\) and \(\frac{7\pi}{4}\). Our plot of the graphs of \(y = \cos(3x)\) and \(y = \cos(5x)\) below (after some careful zooming) bears this out.

Screen Shot 2022-05-17 at 1.06.28 AM.png

  • Unlike the previous problem, there seems to be no quick way to get the circular functions or their arguments to match in the equation \(\sin(x)\cos\left(\frac{x}{2}\right) + \cos(x)\sin\left(\frac{x}{2}\right) = 1\). If we stare at it long enough, however, we realize that the left-hand side is the expanded form of the Sum Identity for \(\sin\left(x + \frac{x}{2}\right)\). Hence, our original equation is equivalent to \(\sin\left(\frac{3}{2} x\right) = 1\). Solving, we find \(x = \frac{\pi}{3} + \frac{4\pi}{3} k\) for integers \(k\). Two of these solutions lie in \([0,2\pi)\): \(x = \frac{\pi}{3}\) and \(x = \frac{5\pi}{3}\). We leave it to the reader to verify the solution by graphing \(y = \sin(x)\cos\left(\frac{x}{2}\right) + \cos(x)\sin\left(\frac{x}{2}\right)\) and \(y = 1\).

9  As always, experience is the greatest teacher here!

10  As always, when in doubt, write it out!

Solving Other Trigonometric Equations

Just as we stated when solving systems of nonlinear equations in Section 9.3  - when it comes to solving equations involving the trigonometric functions, it helps to just try something. There is no way to cover all possible "oddball" cases, but the next example covers two styles that come up often enough.

Example \(\PageIndex{4}\)

Solve the equations and list the solutions which lie in the interval \([0,2\pi)\).

  • \(\cos(x) = \sin(x)\)
  • \(\cos(x) - \sqrt{3} \sin(x) = 2\)
  • This style of problem (along with the next one) often throws students off. The angular frequencies are the same, but the circular functions are different. This has not been an issue up to this point because all the previous examples involving two different circular functions had at least one of these functions being squared. Squares of trigonometric functions are blessings in disguise as we can always try to change them using the Pythagorean Identities. In this case, there are no squares. So what do we do? Since we would like squares, why don't we just square both sides? 11 \[ \begin{array}{lrclr} & \cos(x) & = & \sin(x) & \\ \implies & \cos^2(x) & = & \sin^2(x) & \left( \text{Square both sides} \right) \\ \implies & 1 - \sin^2(x) & = & \sin^2(x) & \left( \text{Pythagorean Identity} \right) \\ \implies & 1 & = & 2 \sin^2(x) &  \\ \implies & \frac{1}{2} & = & \sin^2(x) &  \\ \implies & \pm \frac{1}{\sqrt{2}} & = & \sin(x) &  \\ \end{array} \nonumber \]Therefore, \( \hat{x} = \frac{\pi}{4} \) and we need to check  all  quadrants.\[ \begin{array}{lrclclcl} \text{Solutions associated with QI: } & x & = & \hat{x} + 2\pi k & = & \frac{\pi}{4} + 2 \pi k & & \\ \text{Solutions associated with QII: } & x & = & \pi - \hat{x} + 2\pi k & = & \pi - \frac{\pi}{4} + 2 \pi k & = & \frac{3 \pi}{4} + 2 \pi k \\ \text{Solutions associated with QIII: } & x & = & \pi + \hat{x} + 2\pi k & = & \pi + \frac{\pi}{4} + 2 \pi k & = & \frac{5 \pi}{4} + 2 \pi k \\ \text{Solutions associated with QIV: } & x & = & 2\pi - \hat{x} + 2\pi k & = & 2\pi - \frac{\pi}{4} + 2 \pi k & = & \frac{7 \pi}{4} + 2 \pi k \\ \end{array} \nonumber \]where \( k \in \mathbb{Z} \). This is a case where, rather than compressing these families of solutions into two families (which we  could  do), it's easier to just check the four possibilities separately. In all four cases, the reference angles are \( \hat{x} = \frac{\pi}{4} \). Therefore, the sine and cosine functions become \( \pm \frac{1}{\sqrt{2}} \). We just require the signs of these to be the same. In QI, \( \cos{\left( \frac{\pi}{4} \right)} = \frac{1}{\sqrt{2}} = \sin{\left( \frac{\pi}{4} \right)} \). In QII, \( \cos{\left( \frac{5\pi}{4} \right)} = -\frac{1}{\sqrt{2}} = \sin{\left( \frac{5\pi}{4} \right)} \); however, we can throw out the other two possibilities because the signs of the functions are opposite in both QII and QIV. Thus, the solution set on the interval \( \left[ 0, 2\pi \right) \) is \( x = \frac{\pi}{4} \text{ and } \frac{5 \pi}{4} \).
  • With the absence of double angles or squares, there doesn’t seem to be much we can do. However, since the arguments of the cosine and sine are the same, we can rewrite the left hand side of this equation as a sinusoid. 12  To fit \(f(x) = \cos(x) - \sqrt{3} \sin(x)\) to the form \(A\sin(B t + C) + D\), we use what we learned in Example 10.5.3 and find \(A = 2\), \(D = 0\), \(B = 1\) and \(C = \frac{5\pi}{6}\). Hence, we can rewrite the equation \(\cos(x) - \sqrt{3} \sin(x) = 2\) as \(2 \sin\left(x + \frac{5\pi}{6}\right) = 2\), or \(\sin\left(x + \frac{5\pi}{6}\right) = 1\). Solving the latter, we get \(x = - \frac{\pi}{3} + 2\pi k\) for integers \(k\). Only one of these solutions, \(x = \frac{5\pi}{3}\), which corresponds to \(k=1\), lies in \([0,2\pi)\).

11  As you should recall from Algebra, when you square both sides of an equation, you run the risk of introducing extraneous solutions. Therefore, you must  always  check your solutions when you perform this action.

12  We are essentially "undoing" the Sum and Difference Identity for cosine or sine, depending on which form we use, so this problem is actually closely related to Example \( \PageIndex{3} \) Part e!

Solving Trigonometric Inequalities

Next, we focus on solving inequalities involving the trigonometric functions. Since these functions are continuous on their domains, we may use the sign diagram technique we’ve used in the past to solve the inequalities. 13

Example \( \PageIndex{5} \)

Solve the following inequalities on \([0,2\pi)\). Express your answers using interval notation and verify your answers graphically.

  • \(2\sin(x) \leq 1\)
  • \(\sin(2x) > \cos(x)\)
  • \(\tan(x) \geq 3\)

Screen Shot 2022-05-17 at 1.24.57 AM.png

13  See Example 3.1.5 , Example 6.3.2 , and Example 6.4.2 for discussion of this technique.

14  See  Section 3.1  for a discussion of the non-standard character known as the interrobang.

15  We could have chosen any value \(\arctan (t)\) where \(t>3\).

16  ...by adding \(\pi\) through the inequality...

Computing Domains of Nonstandard Trigonometric Functions

We now provide an example that puts solving equations and inequalities to good use – finding domains of functions.

Example \( \PageIndex{6} \)

Express the domain of the following functions using extended interval notation. 17

  • \(f(x) = \csc\left(2x + \frac{\pi}{3}\right)\)
  • \(f(x) = \dfrac{\sin(x)}{2\cos(x) - 1}\)
  • \(f(x) = \sqrt{1 - \cot(x)}\)

Screen Shot 2022-05-17 at 3.05.46 PM.png

17  See Section 10.3 for details about this notation.

18  This doesn’t necessarily mean the period of \(f\) is \(2 \pi\). The tangent function is comprised of \(\cos (x)\) and \(\sin (x)\), but its period is half theirs. The reader is invited to investigate the period of \(f\).

Solving Equations and Inequalities Involving Inverse Trigonometric Functions

We close this section with an example which demonstrates how to solve equations and inequalities involving the inverse trigonometric functions.

Example \( \PageIndex{7} \)

Solve the following equations and inequalities analytically. Check your answers using a graphing utility.

  • \(\arcsin (2 x)=\frac{\pi}{3}\)
  • \(4 \arccos (x)-3 \pi=0\)
  • \(3 \operatorname{arcsec}(2 x-1)+\pi=2 \pi\)
  • \(4 \arctan ^{2}(x)-3 \pi \arctan (x)-\pi^{2}=0\)
  • \(\pi^{2}-4 \arccos ^{2}(x)<0\)
  • \(4 \operatorname{arccot}(3 x)>\pi\)
  • To solve \(\arcsin (2 x)=\frac{\pi}{3}\), we first note that \(\frac{\pi}{3}\) is in the range of the arcsine function (so a solution exists!) Next, we exploit the inverse property of sine and arcsine from Theorem 10.6.1  \[\begin{aligned} \arcsin (2 x) &=\frac{\pi}{3} \\ \sin (\arcsin (2 x)) &=\sin \left(\frac{\pi}{3}\right) \\ 2 x &=\frac{\sqrt{3}}{2} \quad \text { Since } \sin (\arcsin (u))=u \\ x &=\frac{\sqrt{3}}{4} \end{aligned}\nonumber\] Graphing \(y=\arcsin (2 x)\) and the horizontal line \(y=\frac{\pi}{3}\), we see they intersect at \(\frac{\sqrt{3}}{4} \approx 0.4430\).

Screen Shot 2022-05-17 at 3.34.27 PM.png

  • From \(3 \operatorname{arcsec}(2 x-1)+\pi=2 \pi\), we get \(\operatorname{arcsec}(2 x-1)=\frac{\pi}{3}\). As we saw in Section 10.6 , there are two possible ranges for the arcsecant function. Fortunately, both ranges contain \(\frac{\pi}{3}\). Applying Theorem 10.6.4 , we get \[\begin{aligned} \operatorname{arcsec}(2 x-1) &=\frac{\pi}{3} \\ \sec (\operatorname{arcsec}(2 x-1)) &=\sec \left(\frac{\pi}{3}\right) \\ 2 x-1 &=2 \quad \text { Since } \sec (\operatorname{arcsec}(u))=u \\ x &=\frac{3}{2} \end{aligned}\nonumber\]To check using our calculator, we need to graph \(y=3 \operatorname{arcsec}(2 x-1)+\pi\). To do so, we make use of the identity \(\operatorname{arcsec}(u)=\arccos \left(\frac{1}{u}\right)\) from Theorem 10.6.4 . 19  We see the graph of \(y=3 \arccos \left(\frac{1}{2 x-1}\right)+\pi\) and the horizontal line \(y=2 \pi\) intersect at \(\frac{3}{2}=1.5\).

With the presence of both \(\arctan ^{2}(x)\left(=(\arctan (x))^{2}\right)\) and \(\arctan (x)\), we substitute \(u=\arctan (x)\). The equation \(4 \arctan ^{2}(x)-3 \pi \arctan (x)-\pi^{2}=0\) becomes \(4 u^{2}-3 \pi u-\pi^{2}=0\). Factoring, 20  we get \((4 u+\pi)(u-\pi)=0\), so \(u=\arctan (x)=-\frac{\pi}{4}\) or \(u=\arctan (x)=\pi\). Since \(-\frac{\pi}{4}\) is in the range of arctangent, but \(\pi\) is not, we only get solutions from the first equation. Using Theorem 10.6.2 , we get \[\begin{aligned} \arctan (x) &=-\frac{\pi}{4} \\ \tan (\arctan (x)) &=\tan \left(-\frac{\pi}{4}\right) \\ x &=-1 \quad \text { Since } \tan (\arctan (u))=u \end{aligned}\nonumber\] The calculator verifies our result.

Screen Shot 2022-05-17 at 3.53.39 PM.png

19  Since we are checking for solutions where arcsecant is positive, we know \(u=2 x-1 \geq 1\), and so the identity applies in both cases.

20  It’s not as bad as it looks... don’t let the \(\pi\) throw you!

21  Set \(3x\) equal to the cotangents of the "common angles" and choose accordingly.

  • IIT JEE Study Material

Trigonometric Inequality

An inequality of the standard form R(x) > 0 (or < 0) that consists of 1 or a few trigonometric functions of the variable arc x is a trigonometric inequality . Finding a solution to the inequality means determining the values of the variable arc x, whose trigonometric functions make the inequality true. The solution sets of trigonometric inequalities are expressed in intervals.

The trigonometric unit circle is defined as a circle with a radius of one unit with origin O. The rotation of the variable arc in the counterclockwise direction on the trigonometric unit circle explains 4 common trigonometric functions of the arc, namely

Trigonometric Inequality Examples

The trigonometric unit circle can be used as proof in solving trigonometric inequalities.

The least multiple of all the periods of the trigonometric functions in the inequality is the common period of trigonometric inequality.

The 4 basic types of trigonometric inequalities are as follows:

  • sin x < a (or > a)
  • cos x < a (or > a)
  • tan x < a (or > a)
  • cot x < a (or > a)

Where, a is a given number.

The steps to find the solution for trigonometric inequality are as follows:

1] The inequality is converted into a trigonometric equation by using the equality sign in place of the inequality sign.

2] The trigonometric equation is solved, and solutions are obtained as angle values in the interval [0, 2π].

3] The positive angle greater than π is converted to the equivalent negative value for the fact that the repeated basic interval may lie on the negative side of the origin.

4] The base interval is developed between two values.

5] If the function asymptotes within the interval are developed, then the angle value at which the function asymptotes limits the value of the basic interval.

6] Finally, the solution is generalised.

  • Trigonometric Equations
  • Trigonometric Equations and Its Solutions

Solved Problems on Trigonometric Inequality

Example 1: Solve trigonometric inequality given by sin x ≥ 1 / 2.

The solution of the corresponding equal equation is obtained as:

sin x = 1 / 2 = sin π / 6

⇒ x = π / 6

The sine function is positive in the first and second quarters. Hence, the second angle between “0” and “2π” is:

⇒ x = π − θ = π − π / 6 = 5π / 6

Both angles are less than “π”. Thus, we do not need to convert the angle into an equivalent negative angle. Further, the sine curve is defined for all values of “x”. The base interval, therefore, is π / 6 ≤ x ≤ 5π / 6.

The periodicity of the sine function is “2π”. Hence, we add “2nπ” on either side of the base interval:

2nπ + [π / 6] ≤ x ≤ 2nπ + [5π / 6], n ∈ Z

Example 2: Solve trigonometric inequality given by sinx > cosx.

In order to solve this inequality, it is required to convert it in terms of the inequality of a single trigonometric function.

sinx > cosx

⇒ sinx − cosx > 0

⇒ sin x * cos π / 4 − cos x * sin π / 4 > 0

⇒ sin (x − π / 4) > 0

Let y = x − π/4.

sin y > 0

Thus, we see that problem finally reduces to solving trigonometric sine inequality. The solution of the corresponding equality is obtained as:

⇒ sin y = 0 = sin 0 ⇒ y = 0

The second angle between “0” and “2π” is “π”. The base interval, therefore, is: 0<y<π

2nπ < y < 2nπ + π, n ∈ Z

Now, substituting for y = x − π/4, we have:

2nπ < [x − π / 4] < 2nπ + π, n ∈ Z

⇒ 2nπ + π / 4 < x < 2nπ + 5π / 4, n ∈ Z

Example 3: If the domain of a function, “f(x)”, is [0,1], then find the domain of the function given by f (2 sinx − 1).

The domain of the function is given here. We need to find the domain when the argument (input) to the function is a trigonometric expression. The given domain is: 0≤x≤1

Changing the argument of the function, the domain becomes:

0 ≤ 2 sinx − 1 ≤ 1 ⇒ 1 ≤ 2 sinx ≤ 2 ⇒ ½ ≤ sin x ≤ 1

However, the range of sinx is [-1, 1]. It means that the above interval is equivalent to a trigonometric inequality given by sin x ≥ 1 / 2.

The sine function is positive in the first and second quadrants. Two values of “x” between “0” and “2π” are:

π / 6, π − π / 6

⇒ π / 6, 5π / 6

The value of “x” satisfies the above equation:

2nπ + π/6 < = x < = 2nπ + 5π/6, n ∈ Z

Hence, the required domain is:

Example 4: Solve the inequality tan x < 1.

Hints: Here, the corresponding trigonometric equation is: tanx = 1 = tan π / 4

The other solution in [0, 2π] is x = π + π / 4 = 5π / 4

Corresponding negative angle: y = 5π / 4 − 2π = −3π / 4

However, the tangent function asymptotes at – π/2. Hence, the basic interval is: (−π / 2, π / 4)

Further, the tangent function has a period of π. The general solution is:

nπ − π / 2 < x < nπ + π / 4; n ∈ Z

Example 5: If 12 cot 2 θ – 31 cosec θ + 32 = 0, then find the value of sin θ.

Example 6:   The only value of x for which \(\begin{array}{l}{{2}^{\sin x}}+{{2}^{\cos x}}> {{2}^{1-(1/\sqrt{2})}}\ \text{holds, is}\end{array} \)   

\(\begin{array}{l}A) \frac{5\pi }{4}\\ B) \frac{3\pi }{4}\\ C) \frac{\pi }{2}\end{array} \)

D) All values of x

Solution:  

Since AM ≥ GM

\(\begin{array}{l}\frac{1}{2}({{2}^{\sin x}}+{{2}^{\cos x}})\ge \sqrt{{{2}^{\sin x}}{{.2}^{\cos x}}}\\ \Rightarrow {{2}^{\sin x}}+{{2}^{\cos x}}\ge {{2.2}^{\frac{\sin x+\cos x}{2}}}\\ \Rightarrow {{2}^{\sin x}}+{{2}^{\cos x}}\ge {{2}^{1+\frac{\sin x+\cos x}{2}}}\end{array} \) and we know that 

\(\begin{array}{l}\sin x+\cos x\ge -\sqrt{2} \\ \therefore \ {{2}^{\sin x}}+{{2}^{\cos x}}>{{2}^{1-(1/\sqrt{2})}}, \text \ for \ x=\frac{5\pi }{4}\end{array} \)

Example 7: Common roots of the equations \(\begin{array}{l}2{{\sin }^{2}}x+{{\sin }^{2}}2x=2\ \text{and}\ \sin 2x+\cos 2x=\tan x,\ \text{are}\end{array} \)  

\(\begin{array}{l}A) x=(2n-1)\frac{\pi }{2}\\ B) x=(2n+1)\frac{\pi }{4} \\C) x=(2n+1)\frac{\pi }{3}\end{array} \)

D) None of these

\(\begin{array}{l}2{{\sin }^{2}}x+{{\sin }^{2}}2x=2 \rightarrow (i) \\ \sin 2x+\cos 2x=\tan x \rightarrow (ii)\end{array} \)

Solving (i), 

\(\begin{array}{l}{{\sin }^{2}}2x=2{{\cos }^{2}}x\\ 2{{\cos }^{2}}x\cos 2x=0\\ x=(2n+1)\frac{\pi }{2}\text{ or }x=(2n+1)\frac{\pi }{4}\end{array} \)

Common roots are \(\begin{array}{l}(2n\pm 1)\frac{\pi }{4}\end{array} \)

Solving (ii),

\(\begin{array}{l}\frac{2\tan x+1-{{\tan }^{2}}x}{1+{{\tan }^{2}}x}=\tan x\\ \Rightarrow {{\tan }^{3}}x+{{\tan }^{2}}x-\tan x-1=0\\ \Rightarrow ({{\tan }^{2}}x-1)\,(\tan x+1)=0\\ \Rightarrow x=m\pi \pm \frac{\pi }{4}\end{array} \)

Example 8: If sin θ = √3 cos θ, -π < θ < 0, then find θ.

\(\begin{array}{l}\tan \theta =\sqrt{3}=\tan \frac{\pi }{3}\Rightarrow \theta =n\pi +\frac{\pi }{3}\end{array} \)

For -π < θ < 0

Put n = −1, we get 

Frequently Asked Questions

What do you mean by trigonometric inequality.

Trigonometry inequality is an inequality that has one or more trigonometry functions in the form of R [f(x), g(x)…] > 0 (or < 0), where f(x), g(x),…etc., are trigonometric functions of x.

Give an example of trigonometric inequality.

An example of trigonometric inequality: sin x + sin 3x < 1.

Give the solution of the inequality of the form tan x > a for any real value of a.

The solution of the inequality tan x > a, (for any real value of a) has the form arctan a + πn < x < π/2 + πn, n ∈ Z.

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Unit 2: Solving equations & inequalities

About this unit, linear equations with variables on both sides.

  • Why we do the same thing to both sides: Variable on both sides (Opens a modal)
  • Intro to equations with variables on both sides (Opens a modal)
  • Equations with variables on both sides: 20-7x=6x-6 (Opens a modal)
  • Equation with variables on both sides: fractions (Opens a modal)
  • Equation with the variable in the denominator (Opens a modal)
  • Equations with variables on both sides Get 3 of 4 questions to level up!
  • Equations with variables on both sides: decimals & fractions Get 3 of 4 questions to level up!

Linear equations with parentheses

  • Equations with parentheses (Opens a modal)
  • Reasoning with linear equations (Opens a modal)
  • Multi-step equations review (Opens a modal)
  • Equations with parentheses Get 3 of 4 questions to level up!
  • Equations with parentheses: decimals & fractions Get 3 of 4 questions to level up!
  • Reasoning with linear equations Get 3 of 4 questions to level up!

Analyzing the number of solutions to linear equations

  • Number of solutions to equations (Opens a modal)
  • Worked example: number of solutions to equations (Opens a modal)
  • Creating an equation with no solutions (Opens a modal)
  • Creating an equation with infinitely many solutions (Opens a modal)
  • Number of solutions to equations Get 3 of 4 questions to level up!
  • Number of solutions to equations challenge Get 3 of 4 questions to level up!

Linear equations with unknown coefficients

  • Linear equations with unknown coefficients (Opens a modal)
  • Why is algebra important to learn? (Opens a modal)
  • Linear equations with unknown coefficients Get 3 of 4 questions to level up!

Multi-step inequalities

  • Inequalities with variables on both sides (Opens a modal)
  • Inequalities with variables on both sides (with parentheses) (Opens a modal)
  • Multi-step inequalities (Opens a modal)
  • Using inequalities to solve problems (Opens a modal)
  • Multi-step linear inequalities Get 3 of 4 questions to level up!
  • Using inequalities to solve problems Get 3 of 4 questions to level up!

Compound inequalities

  • Compound inequalities: OR (Opens a modal)
  • Compound inequalities: AND (Opens a modal)
  • A compound inequality with no solution (Opens a modal)
  • Double inequalities (Opens a modal)
  • Compound inequalities examples (Opens a modal)
  • Compound inequalities review (Opens a modal)
  • Solving equations & inequalities: FAQ (Opens a modal)
  • Compound inequalities Get 3 of 4 questions to level up!

IMAGES

  1. Solving Trig Equations 2

    how to solve trig equations and inequalities

  2. Math Exercises & Math Problems: Trigonometric Equations and Inequalities

    how to solve trig equations and inequalities

  3. Solving trig equations

    how to solve trig equations and inequalities

  4. solving trig equations on an interval

    how to solve trig equations and inequalities

  5. Solving Trig Equations

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  6. Trigonometric Inequalities Steps

    how to solve trig equations and inequalities

VIDEO

  1. 5.3A Solve trig equations

  2. AP PreCal 3.10 Trig Inequalities (video 2)

  3. Section 10.7

  4. Section 10.7

  5. Precalculus BC Unit 9 Notes 5 Solving Trig Inequalities and More Equations

  6. Trigonometry || IIT&JEE Questions NO 03|| X Class

COMMENTS

  1. 10.7: Trigonometric Equations and Inequalities

    Since the argument of sine here is 3x, we have 3x = π 6 + 2πk or 3x = 5π 6 + 2πk for integers k. To solve for x, we divide both sides 2 of these equations by 3, and obtain x = π 18 + 2π 3 k or x = 5π 18 + 2π 3 k for integers k. This is the technique employed in the example below. Example 10.7.1.

  2. Trigonometric equations and identities

    Quiz Unit test About this unit 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.

  3. 100 Fully Solved Trigonometric Equations and Inequalities

    http://www.greenemath.com/Step-by-Step full solutions for our practice tests on solving trigonometric equations and inequalities. These problems will cover: ...

  4. Calculus I

    Most trig equations won't come down to one of those and will in fact need a calculator to solve. The next section is devoted to this kind of problem. In this section we will discuss how to solve trig equations. The answers to the equations in this section will all be one of the "standard" angles that most students have memorized after a trig class.

  5. Trigonometric Inequalities Calculator

    To solve trigonometric inequalities, use the properties of the trigonometric functions and algebraic manipulations to isolate the variable on one side of the inequality. Then, apply appropriate trigonometric function inverses to find the interval. What are trigonometric inequalities?

  6. Solving Trigonometric Equations and Inequalities

    Possible Answers: Use the trigonometric identities to switch sec into terms of tan: for n being any integer. Which of the following is not a solution to

  7. Algebra Trig Review

    Solve each of the following inequalities. Show All Solutions Hide All Solutions x2−10 > 3x x 2 − 10 > 3 x Show Solution x4+4x3−12x2 ≤ 0 x 4 + 4 x 3 − 12 x 2 ≤ 0 Show Solution 3x2 −2x−11 > 0 3 x 2 − 2 x − 11 > 0 Show Solution x −3 x +2 ≥ 0 x − 3 x + 2 ≥ 0 Show Solution x2 −3x−10 x−1 < 0 x 2 − 3 x − 10 x − 1 < 0 Show Solution

  8. 3.10 Trigonometric Equations and Inequalities

    trigonometric inequalities and equations. Be sure to pay attention to whether you need to use degree mode or radian mode to find the correct answer. Solving Trigonometric Equations It's important to note that when using inverse trigonometric functions , the solutions you find may need to be modified due to domain restrictions .

  9. Solving Trigonometric Equations and Inequalities Full Course

    http://www.greenemath.com/In this lesson, we will learn how to solve the various types of trigonometric equations and inequalities. We will cover inverse tri...

  10. How to Solve Trigonometric Equations: A Simple Tutorial

    1 Know the Solving concept. [1] To solve a trig equation, transform it into one or many basic trig equations. Solving trig equations finally results in solving 4 types of basic trig equations. 2 Know how to solve basic trig equations. [2] There are 4 types of basic trig equations: sin x = a ; cos x = a tan x = a ; cot x = a

  11. PDF SOLVING TRIGONOMETRIC INEQUALITIES

    Solving basic trig inequalities proceeds by using trig conversion tables (or calculators), then by considering the various positions of the variable arc x that rotates on the trig circle. Example 1. Solve the inequality: sin x > 0.709 Solution. The solution set is given by both trig table and trig unit circle.

  12. Section 2.4 : Solving Trig Equations

    Solve the following trig equations. For those without intervals listed find ALL possible solutions. For those with intervals listed find only the solutions that fall in those intervals. Show All Solutions Hide All Solutions. 2cos(t) =√3 2 cos ( t) = 3.

  13. Trigonometric Equation Calculator

    To solve a trigonometric simplify the equation using trigonometric identities. Then, write the equation in a standard form, and isolate the variable using algebraic manipulation to solve for the variable. Use inverse trigonometric functions to find the solutions, and check for extraneous solutions. What is a basic trigonometric equation?

  14. Solving Trigonometric Equations Using Identities, Multiple Angles, By

    This trigonometry video tutorial shows you how to solve trigonometric equations using identities with multiple angles, by factoring, and by finding the gener...

  15. 10.7: Trigonometric Equations and Inequalities

    We will repeat this technique in the example below. Example 10.7.1. Solve the following equations and check your answers analytically. List the solutions which lie in the interval [0, 2π) and verify them analytically (i.e., by hand) or by using graphing technology. sin(2x) = − √3 2. csc(1 3x − π) = √2. cot(3x) = 0.

  16. Trigonometric Inequality

    Trigonometric Equations and Its Solutions Solved Problems on Trigonometric Inequality Example 1: Solve trigonometric inequality given by sin x ≥ 1 / 2. Solution: The solution of the corresponding equal equation is obtained as: sin x = 1 / 2 = sin π / 6 ⇒ x = π / 6 The sine function is positive in the first and second quarters.

  17. 3.10 Trig Equations and Inequalities

    3.10.A Solve equations and inequalities involving trigonometric functions. *AP® is a trademark registered and owned by the CollegeBoard, which was not involved in the production of, and does not endorse, this site.

  18. Solving equations & inequalities

    Unit test. Level up on all the skills in this unit and collect up to 1100 Mastery points! There are lots of strategies we can use to solve equations. Let's explore some different ways to solve equations and inequalities. We'll also see what it takes for an equation to have no solution, or infinite solutions.

  19. Solving Trigonometric Equations

    Solving trigonometric equations - from the basics to more challenging problems. This is a large topic but with practice and a good understanding of the funda...

  20. PDF I Solve Each Equation in The Indicated Interval All Angles Are to Be

    Dr. Little's equations), but if you are able to solve these, you will be in good shape for the exam. SOLVE ALL EQUAITONS IN RADIANS IN THE INTERVALS INDICATED. 1. 7x IN )S 2. 22 x SOLVE IN THE INTERVAL 3. 3x IN ] SS 4. sin( ) 1x IN 5. 22 x IN Note: 3 424 6. 9x IN [ , ] SS 7. 7x SOLVE IN THE INTERVAL )f 8. 7x IN 9. 0x IN 10.

  21. Calculus I

    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 ( 3 t) = 2 in [0, 4π 3] [ 0, 4 π 3] Solution 2cos( x 3) +√2 = 0 2 cos

  22. ChatSAT on Instagram: " Mastering the SAT is all about strategy!

    11 likes, 0 comments - chatsat.io on February 10, 2024: " Mastering the SAT is all about strategy! Here's your roadmap to success: ⁣ Av..."

  23. Eye Level Bandar Puchong Jaya on Instagram: " Explore comprehensive

    2 likes, 0 comments - eyelevelbdrpuchongjaya on February 14, 2024: " Explore comprehensive secondary school math education with Eye Level Math at Eye Level Banda..."

  24. Algebra

    In this chapter we will look at one of the standard topics in any Algebra class. The ability to solve equations and/or inequalities is very important and is used time and again both in this class and in later classes.

  25. 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. 10cos(8t) =−5 10 cos. ⁡. ( 8 t) = − 5. 10cos(8t) =−5 10 cos. ⁡.