solving trig equations by factoring

• school Campus Bookshelves
• menu_book Bookshelves
• perm_media Learning Objects
• login Login
• how_to_reg Request Instructor Account
• hub Instructor Commons
• Download Page (PDF)
• Download Full Book (PDF)
• Periodic Table
• Physics Constants
• Scientific Calculator
• Reference & Cite
• Tools expand_more
• Readability

selected template will load here

This action is not available.

3.2.8: Trigonometric Equations Using Factoring

• Last updated
• Save as PDF
• Page ID 14832

Factoring and the Quadratic Formula.

Solving trig equations is an important process in mathematics. Quite often you'll see powers of trigonometric functions and be asked to solve for the values of the variable which make the equation true. For example, suppose you were given the trig equation

\(2\sin x \cos x=\cos x\)

Trigonometric Equations Using Factoring

You have no doubt had experience with factoring. You have probably factored equations when looking for the possible values of some variable, such as "\(x\)". It might interest you to find out that you can use the same factoring method for more than just a variable that is a number. You can factor trigonometric equations to find the possible values the function can take to satisfy an equation.

Algebraic skills like factoring and substitution that are used to solve various equations are very useful when solving trigonometric equations . As with algebraic expressions, one must be careful to avoid dividing by zero during these maneuvers.

Solving for Unknown Values

1. Solve \(2\sin^2 x−3\sin x+1=0\) for \(0<x \leq 2\pi\).

\(\begin{aligned} \qquad \qquad \quad x^2 \sin^2 x−3\sin x+1&=0 \qquad \text{Factor this like a quadratic equation} \\ \qquad \qquad \quad (2\sin x−1)(\sin x−1)&=0 \end{aligned} \\ \begin{aligned} & \downarrow & \searrow& \\ 2\sin x−1&=0 &\text{or} \quad \sin x−1&=0 \\ 2\sin x&=1 & \sin x&=1 \\ \sin x&=\dfrac{1}{2} & x&=\dfrac{\pi}{2} \\ x=\dfrac{\pi}{6} \text{ and } x&=\dfrac{5\pi}{6} & & \end{aligned} \)

2. Solve \(2 \tan x \sin x+2\sin x=\tan x+1\) for all values of \(x\).

Pull out \(\sin x\)

There is a common factor of \((\tan x+1)\)

Think of the \(−(\tan x+1)\) as \((−1)(\tan x+1)\), which is why there is a \(−1\) behind the \(2 \sin x\).

3. Solve \(2\sin^2 x+3\sin x−2=0\) for all \(x\), \([0,\pi ]\).

\(\begin{aligned} \qquad \quad 2 \sin ^{2} x+3 \sin x-2&=0 \rightarrow \text { Factor like a quadratic }\\ \qquad \quad (2 \sin x-1)(\sin x+2)&=0 \end{aligned} \\ \begin{aligned} \swarrow&& \qquad \searrow &\\ 2 \sin x-1 &=0 & \sin x+2&=0 \\ \sin x&=\dfrac{1}{2} & \sin x&=-2 \end{aligned}\) \(x=\dfrac{\pi}{6} \text { and } x=\dfrac{5 \pi}{6} \text { There is no solution because the range of } \sin x \text { is }[-1,1] \text { . }\)

Some trigonometric equations have no solutions. This means that there is no replacement for the variable that will result in a true expression.

Example \(\PageIndex{1}\)

Earlier, you were asked to solve this:

\(2 \sin x \cos x=\cos x\)

Subtract \(\cos x\) from both sides and factor it out of the equation:

\(\begin{aligned} 2\sin x \cos x−\cos x&=0 \\ \cos x(2\sin x−1)&=0 \end{aligned}\)

Now set each factor equal to zero and solve. The first is \(\cos x\):

\(\begin{aligned} \cos x&=0 \\ x&=\dfrac{\pi}{2},\; \dfrac{3\pi }{2}\end{aligned}\)

And now for the other term:

\(\begin{aligned} 2 \sin x−1&=0 \\ \sin x&=\dfrac{1}{2} \\ x&=\dfrac{\pi}{6},\; \dfrac{5\pi}{6}\end{aligned}\)

Example \(\PageIndex{2}\)

Solve the trigonometric equation \(4 \sin x \cos x+2\cos x−2\sin x−1=0\) such that \(0\leq x<2\pi\).

Use factoring by grouping.

\(\begin{aligned} 2 \sin x+1&=0 & \text { or } \qquad 2 \cos x-1&=0 \\ 2 \sin x&=-1 & 2 \cos x&=1 \\ \sin x&=-\dfrac{1}{2} & \cos x&=\dfrac{1}{2} \\ x&=\dfrac{7 \pi}{6},\; \dfrac{11 \pi}{6} & x&=\dfrac{\pi}{3},\; \dfrac{5 \pi}{3} \end{aligned}\)

Example \(\PageIndex{3}\)

Solve \(\tan^2 x=3\tan x\) for \(x\) over \([0,\pi ]\).

\(\begin{aligned} \tan ^{2} x &=3 \tan x & & &\\ \tan ^{2} x-3 \tan x &=0 & & &\\ \tan x(\tan x-3) &=0 & & &\\ \tan x &=0 \qquad \text { or } \quad &\tan x&=3 \\ x &=0, \pi & x&=1.25 \end{aligned}\)

Example \(\PageIndex{4}\)

Find all the solutions for the trigonometric equation \(2 \sin^2\dfrac{x}{4}−3\cos \dfrac{x}{4}=0\) over the interval \([0,2\pi )\).

\(2\sin^2 \dfrac{x}{4}−3\cos \dfrac{x}{4}=0\)

\(\begin{array}{c} &\qquad \qquad \quad 2\left(1-\cos ^{2} \dfrac{x}{4}\right)-3 \cos \dfrac{x}{4}=0 \\ & \qquad \qquad \quad 2-2 \cos ^{2} \dfrac{x}{4}-3 \cos \dfrac{x}{4}=0 \\ & \qquad \qquad \quad 2 \cos ^{2} \dfrac{x}{4}+3 \cos \dfrac{x}{4}-2=0 \\ &\qquad \qquad \quad \left(2 \cos \dfrac{x}{4}-1\right)\left(\cos \dfrac{x}{4}+2\right)=0 \end{array} \\ \begin{aligned} &\swarrow & \searrow &\\ 2 \cos \dfrac{x}{4}-1&=0 &\text { or } \quad \cos \dfrac{x}{4}+2&=0 \\ 2 \cos \dfrac{x}{4}&=1 & \cos \dfrac{x}{4}&=-2 \\ \cos \dfrac{x}{4}&=\dfrac{1}{2} & & \\ \dfrac{x}{4}=\dfrac{\pi}{3} \quad \text { or } \quad& \dfrac{5 \pi}{3} && \\ x=\dfrac{4 \pi}{3} \quad \text { or }\quad& \dfrac{20 \pi}{3} && \end{aligned}\)

\(\dfrac{20\pi}{3}\) is eliminated as a solution because it is outside of the range and \(\cos x^4=−2\) will not generate any solutions because \(−2\) is outside of the range of cosine. Therefore, the only solution is \(\dfrac{4\pi}{3}\).

Solve each equation for \(x\) over the interval \([0,2\pi )\).

• \(\cos^2 (x)+2 \cos(x)+1=0\)
• \(1−2\sin (x)+\sin^2 (x)=0\)
• \(2\cos (x) \sin (x)−\cos(x)=0\)
• \(\sin (x) \tan^2 (x)−\sin (x) =0\)
• \(\sec^2 (x)=4\)
• \(\sin^2 (x)−2\sin (x) =0\)
• \(3\sin (x) =2\cos^2 (x)\)
• \(2\sin^2 (x)+3\sin (x) =2\)
• \(\tan(x) \sin^2 (x)=\tan(x)\)
• \(2\sin^2 (x)+\sin (x) =1\)
• \(2\cos(x)\tan(x)−\tan (x)=0\)
• \(\sin^2 (x)+\sin (x) =2\)
• \(\tan(x)(2 \cos^2 (x)+3\cos(x)−2)=0\)
• \(\sin^2 (x)+1=2\sin (x)\)
• \(2\cos^2 (x)−3\cos (x)=2\)

Review (Answers)

To see the Review answers, open this PDF file and look for section 3.4.

Additional Resources

Video: Example: Solve a Trig Equation by Factoring

Practice: Trigonometric Equations Using Factoring

Precalculus

Trigonometry, solving trigonometric equations by factoring.

In many cases, the left side of the trigonometric equation

can be factored. If f ( x ) can be represented as

then the solution of the original equation will be all the roots of the equations

which belong to the domain of the original equation f ( x ) = 0 .

Factoring can be done in various ways: by the grouping method, using short multiplication formulas, etc.

For quadratic expressions , the following factoring formula is true:

where \(x_1\) and \(x_2\) are the roots of the equation

For example, the trigonometric equation

contains a quadratic expression whose roots are \({\frac{1}{2}}\) and \(1.\) In this case, this equation is represented in the form

As you can see, the problem is reduced to solving two basic trigonometric equations :

Solved Problems

Solve the equation

Note that \(\cos x = 1\) is a solution to the equation. Therefore, we can group and factor the left-hand side of the equation as follows:

Equating to zero each of the factors on the left side of the equation, we get:

Let \(t = \cos x.\) The quadratic equation \(4t^2 +4t + 7 = 0\) has no solutions since its discriminant is negative:

Solve the trigonometric equation

Using the sum of cosines identity we can write the first two terms as

Now we can factor the left side of the equation:

We get two basic equations :

It's obvious that

We can factor the left-hand side:

Solve the two equations:

Make sure that the root \(x_1\) satisfies the domain of the equation:

It can be seen that points \(x_1 = \frac{\pi n}{3}\) do not intersect with points \(x = \frac{\pi}{6} + \frac{\pi n}{3}.\)

We transform the left side of the equation into product using the sum of sines formula :

We take into account that the cosine function is even. Then we have

By the double-angle identity \(\sin 2x = 2\sin x \cos x.\) This yields:

The original equation was reduced to solving two elementary equations:

HIGH SCHOOL

• ACT Tutoring
• SAT Tutoring
• PSAT Tutoring
• ASPIRE Tutoring
• SHSAT Tutoring
• STAAR Tutoring

GRADUATE SCHOOL

• MCAT Tutoring
• GRE Tutoring
• LSAT Tutoring
• GMAT Tutoring
• AIMS Tutoring
• HSPT Tutoring
• ISAT Tutoring
• SSAT Tutoring

Search 50+ Tests

Loading Page

math tutoring

• Elementary Math
• Pre-Calculus
• Trigonometry

science tutoring

Foreign languages.

• Mandarin Chinese

elementary tutoring

• Computer Science

Search 350+ Subjects

• Video Overview
• Tutor Selection Process
• Online Tutoring
• Mobile Tutoring
• Instant Tutoring
• How We Operate
• Our Guarantee
• Impact of Tutoring
• Reviews & Testimonials
• Media Coverage
• About Varsity Tutors

Trigonometry : Factoring Trigonometric Equations

Study concepts, example questions & explanations for trigonometry, all trigonometry resources, example questions, example question #1 : factoring trigonometric equations.

Example Question #2 : Factoring Trigonometric Equations

1, 2, and 3

Our first choice is valid.

Therefore, our third answer choice is not correct, meaning only 1 is correct.

Find the zeros of the above equation in the interval

Therefore, 

Example Question #4 : Factoring Trigonometric Equations

Factor the expression

This last expression can be written as :

This shows the required result.

Example Question #5 : Factoring Trigonometric Equations

Factor the following expression:

We know that we can write

This is the result that we need.

Example Question #6 : Factoring Trigonometric Equations

We accept that :

What is a simple expression of

First we see that :

We know that :

We can't factor this expression.

Note first that:

We therefore have :

Example Question #8 : Factoring Trigonometric Equations

Factor the following expression

We cannot factor the above expression.

This shows that we cannot factor the above expression.

Example Question #9 : Factoring Trigonometric Equations

We first note that we have:

Example Question #10 : Factoring Trigonometric Equations

Find a simple expression for the following :

First of all we know that :

we have then:

Report an issue with this question

If you've found an issue with this question, please let us know. With the help of the community we can continue to improve our educational resources.

DMCA Complaint

If you believe that content available by means of the Website (as defined in our Terms of Service) infringes one or more of your copyrights, please notify us by providing a written notice (“Infringement Notice”) containing the information described below to the designated agent listed below. If Varsity Tutors takes action in response to an Infringement Notice, it will make a good faith attempt to contact the party that made such content available by means of the most recent email address, if any, provided by such party to Varsity Tutors.

Your Infringement Notice may be forwarded to the party that made the content available or to third parties such as ChillingEffects.org.

Please be advised that you will be liable for damages (including costs and attorneys’ fees) if you materially misrepresent that a product or activity is infringing your copyrights. Thus, if you are not sure content located on or linked-to by the Website infringes your copyright, you should consider first contacting an attorney.

Please follow these steps to file a notice:

You must include the following:

A physical or electronic signature of the copyright owner or a person authorized to act on their behalf; An identification of the copyright claimed to have been infringed; A description of the nature and exact location of the content that you claim to infringe your copyright, in \ sufficient detail to permit Varsity Tutors to find and positively identify that content; for example we require a link to the specific question (not just the name of the question) that contains the content and a description of which specific portion of the question – an image, a link, the text, etc – your complaint refers to; Your name, address, telephone number and email address; and A statement by you: (a) that you believe in good faith that the use of the content that you claim to infringe your copyright is not authorized by law, or by the copyright owner or such owner’s agent; (b) that all of the information contained in your Infringement Notice is accurate, and (c) under penalty of perjury, that you are either the copyright owner or a person authorized to act on their behalf.

Send your complaint to our designated agent at:

Charles Cohn Varsity Tutors LLC 101 S. Hanley Rd, Suite 300 St. Louis, MO 63105

Or fill out the form below:

Contact Information

Complaint details.

• school Campus Bookshelves
• menu_book Bookshelves
• perm_media Learning Objects
• login Login
• how_to_reg Request Instructor Account
• hub Instructor Commons
• Download Page (PDF)
• Download Full Book (PDF)
• Periodic Table
• Physics Constants
• Scientific Calculator
• Reference & Cite
• Tools expand_more
• Readability

selected template will load here

This action is not available.

3.6: Solving Trigonometric Equations

• Last updated
• Save as PDF
• Page ID 60926

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.

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 \(\PageIndex{2}\).

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.

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 \(\PageIndex{4}\).

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 \(\PageIndex{5}\).

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 \(\PageIndex{18}\).

If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

To log in and use all the features of Khan Academy, please enable JavaScript in your browser.

Unit 4: Trigonometric equations and identities

About this unit, inverse trigonometric functions.

• Intro to arcsine (Opens a modal)
• Intro to arctangent (Opens a modal)
• Intro to arccosine (Opens a modal)
• Restricting domains of functions to make them invertible (Opens a modal)
• Domain & range of inverse tangent function (Opens a modal)
• Using inverse trig functions with a calculator (Opens a modal)
• Inverse trigonometric functions review (Opens a modal)
• Trigonometric equations and identities: FAQ (Opens a modal)
• Evaluate inverse trig functions Get 3 of 4 questions to level up!

Sinusoidal equations

• Solving sinusoidal equations of the form sin(x)=d (Opens a modal)
• Cosine equation algebraic solution set (Opens a modal)
• Cosine equation solution set in an interval (Opens a modal)
• Sine equation algebraic solution set (Opens a modal)
• Solving cos(θ)=1 and cos(θ)=-1 (Opens a modal)
• Solve sinusoidal equations (basic) Get 3 of 4 questions to level up!
• Solve sinusoidal equations Get 3 of 4 questions to level up!

Sinusoidal models

• Interpreting solutions of trigonometric equations (Opens a modal)
• Trig word problem: solving for temperature (Opens a modal)
• Trigonometric equations review (Opens a modal)
• Interpret solutions of trigonometric equations in context Get 3 of 4 questions to level up!
• Sinusoidal models word problems Get 3 of 4 questions to level up!

Angle addition identities

• Trig angle addition identities (Opens a modal)
• Using the cosine angle addition identity (Opens a modal)
• Using the cosine double-angle identity (Opens a modal)
• Proof of the sine angle addition identity (Opens a modal)
• Proof of the cosine angle addition identity (Opens a modal)
• Proof of the tangent angle sum and difference identities (Opens a modal)
• Using the trig angle addition identities Get 3 of 4 questions to level up!

Using trigonometric identities

• Finding trig values using angle addition identities (Opens a modal)
• Using the tangent angle addition identity (Opens a modal)
• Using trig angle addition identities: finding side lengths (Opens a modal)
• Using trig angle addition identities: manipulating expressions (Opens a modal)
• Using trigonometric identities (Opens a modal)
• Trig identity reference (Opens a modal)
• Find trig values using angle addition identities Get 3 of 4 questions to level up!

Challenging trigonometry problems

• Trig challenge problem: area of a triangle (Opens a modal)
• Trig challenge problem: area of a hexagon (Opens a modal)
• Trig challenge problem: cosine of angle-sum (Opens a modal)
• Trig challenge problem: arithmetic progression (Opens a modal)
• Trig challenge problem: maximum value (Opens a modal)
• Trig challenge problem: multiple constraints (Opens a modal)
• Trig challenge problem: system of equations (Opens a modal)

Solver Title

Generating PDF...

• Pre Algebra Order of Operations Factors & Primes Fractions Long Arithmetic Decimals Exponents & Radicals Ratios & Proportions Percent Modulo Number Line Mean, Median & Mode
• Algebra Equations Inequalities System of Equations System of Inequalities Basic Operations Algebraic Properties Partial Fractions Polynomials Rational Expressions Sequences Power Sums Interval Notation Pi (Product) Notation Induction Logical Sets Word Problems
• Pre Calculus Equations Inequalities Scientific Calculator Scientific Notation Arithmetics Complex Numbers Polar/Cartesian Simultaneous Equations System of Inequalities Polynomials Rationales Functions Arithmetic & Comp. Coordinate Geometry Plane Geometry Solid Geometry Conic Sections Trigonometry
• Calculus Derivatives Derivative Applications Limits Integrals Integral Applications Integral Approximation Series ODE Multivariable Calculus Laplace Transform Taylor/Maclaurin Series Fourier Series Fourier Transform
• Functions Line Equations Functions Arithmetic & Comp. Conic Sections Transformation
• Linear Algebra Matrices Vectors
• Trigonometry Identities Proving Identities Trig Equations Trig Inequalities Evaluate Functions Simplify
• Statistics Mean Geometric Mean Quadratic Mean Average Median Mode Order Minimum Maximum Probability Mid-Range Range Standard Deviation Variance Lower Quartile Upper Quartile Interquartile Range Midhinge Standard Normal Distribution
• Physics Mechanics
• Chemistry Chemical Reactions Chemical Properties
• Finance Simple Interest Compound Interest Present Value Future Value
• Economics Point of Diminishing Return
• Conversions Roman Numerals Radical to Exponent Exponent to Radical To Fraction To Decimal To Mixed Number To Improper Fraction Radians to Degrees Degrees to Radians Hexadecimal Scientific Notation Distance Weight Time Volume
• Pre Algebra
• One-Step Addition
• One-Step Subtraction
• One-Step Multiplication
• One-Step Division
• One-Step Decimals
• Two-Step Integers
• Two-Step Add/Subtract
• Two-Step Multiply/Divide
• Two-Step Fractions
• Two-Step Decimals
• Multi-Step Integers
• Multi-Step with Parentheses
• Multi-Step Rational
• Multi-Step Fractions
• Multi-Step Decimals
• Solve by Factoring
• Completing the Square
• Quadratic Formula
• Biquadratic
• Logarithmic
• Exponential
• Rational Roots
• Floor/Ceiling
• Equation Given Roots
• Newton Raphson
• Substitution
• Elimination
• Cramer's Rule
• Gaussian Elimination
• System of Inequalities
• Perfect Squares
• Difference of Squares
• Difference of Cubes
• Sum of Cubes
• Polynomials
• Distributive Property
• FOIL method
• Perfect Cubes
• Binomial Expansion
• Negative Rule
• Product Rule
• Quotient Rule
• Expand Power Rule
• Fraction Exponent
• Exponent Rules
• Exponential Form
• Logarithmic Form
• Absolute Value
• Rational Number
• Powers of i
• Partial Fractions
• Is Polynomial
• Leading Coefficient
• Leading Term
• Standard Form
• Complete the Square
• Synthetic Division
• Linear Factors
• Rationalize Denominator
• Rationalize Numerator
• Identify Type
• Convergence
• Interval Notation
• Pi (Product) Notation
• Boolean Algebra
• Truth Table
• Mutual Exclusive
• Cardinality
• Caretesian Product
• Age Problems
• Distance Problems
• Cost Problems
• Investment Problems
• Number Problems
• Percent Problems
• Addition/Subtraction
• Multiplication/Division
• Dice Problems
• Coin Problems
• Card Problems
• Pre Calculus
• Linear Algebra
• Trigonometry
• Conversions

Most Used Actions

Number line.

• factor\:x^{2}-5x+6
• factor\:(x-2)^2-9
• factor\:2x^2-18
• factor\:5a^2-30a+45
• factor\:x^6-2x^4-x^2+2
• factor\:2x^5+x^4-2x-1
• What is the sum of cubes formula?
• The sum of cubes formula is a³ + b³ = (a+b)(a² - ab + b²)
• What is the difference of squares formula?
• The difference of squares formula is a² - b² = (a+b)(a-b)
• What is the difference of cubes formula?
• The difference of cubes formula is a³ - b³ = (a-b)(a² + ab + b²)
• How do you solve factoring by greatest common monomial factor?
• To factor by greatest common monomial factor, find the greatest common monomial factor among the terms of the expression and then factor it out of each term.
• How do you factor a monomial?
• To factor a monomial, write it as the product of its factors and then divide each term by any common factors to obtain the fully-factored form.
• How do you factor a binomial?
• To factor a binomial, write it as the sum or difference of two squares or as the difference of two cubes.
• How do you factor a trinomial?
• To factor a trinomial x^2+bx+c find two numbers u, v that multiply to give c and add to b. Rewrite the trinomial as the product of two binomials (x-u)(x-v)
• How to find LCM with the listing multiples method?
• To find the LCM of two numbers using the listing multiples method write down the multiples of the first number and write down the multiples of the second number. Find the smallest number that is a multiple of both of the numbers.

factor-calculator

• Middle School Math Solutions – Polynomials Calculator, Factoring Quadratics Just like numbers have factors (2×3=6), expressions have factors ((x+2)(x+3)=x^2+5x+6). Factoring is the process...

Please add a message.

Message received. Thanks for the feedback.

• calculators
• Trigonometric Equations

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

 Final Answer

Struggling with math.

Access detailed step by step solutions to thousands of problems, growing every day!

 Popular problems