Solving Trigonometric Equations

This section covers:

Solving trig equations is just finding the solutions of equations like we did with linear, quadratic, and radical equations, but using trig functions instead. We will mainly use the Unit Circle to find the exact solutions if we can, and we’ll start out by finding the solutions from \(\left[ 0,\,\,2\pi \right)\).

We can also solve these using a Graphing Calculator, as we’ll see below.

Note that we will use Trigonometric Identities to solve trig problems in the Trigonometric Identity section.

Important Note: There is a subtle distinction between finding inverse trig functions and solving for trig functions. If we want \(\displaystyle {{\sin }^{{-1}}}\left( {\frac{{\sqrt{2}}}{2}} \right)\) for example, like in the The Inverse Trigonometric Functions section, we only pick the answers from Quadrants I and IV, so we get  \(\displaystyle \frac{\pi }{4}\) only. But if we are solving \(\displaystyle \sin \left( x \right)=\frac{{\sqrt{2}}}{2}\) we get \(\displaystyle \frac{\pi }{4}\) and \(\displaystyle \frac{{3\pi }}{4}\)  in the interval \(\left( {0,2\pi } \right)\); there are no domain restrictions.

Solving Trigonometric Equations Using the Unit Circle

Let’s start out with solving fairly simple Trig Equations and getting the solutions from  \(\left[ 0,\,\,2\pi \right)\), or \(\left[ {0,{{{360}}^{o}}} \right]\).

Here is the Unit Circle again so we can “pick off” the answers from it:


Notice how sometimes we have to divide up the equation into two separate equations, like when the argument of the trig function is an expression, like  \(\displaystyle \theta +\frac{\pi }{{18}}\). Also note that  \({{\left( \cos \theta \right)}^{2}}\)  is written as  \({{\cos }^{2}}\theta \), and we can put it in the graphing calculator as \(\boldsymbol{\cos {{\left( x \right)}^{2}}}\) or \(\boldsymbol {{{\left( {\cos \left( x \right)} \right)}^{2}}}\).

If the coefficient before the θ is less than 1, we may have to “throw away” an extraneous solution (like in the last tan problem below). If we have a coefficient before the θ that is greater than 1, we have to first find the general solution of the equation, and then go back to the Unit Circle to see where the solutions are in the [0, 2π) interval. We will learn how to do this here.


Note that sometimes you may have to solve using degrees [0, 360°) instead of radians. Also note that sometimes we have to divide a sin by a cos to get a tan, as in one of the examples. And the last problem involves solving a trig inequality.

Solving Trig Problems 0 to 2pi

Solving Trigonometric Equations – General Solutions

Since trig functions go on and on in both directions of the \(x\)-axis, we’ll also have to know how to solve trig equations over the set of real numbers; this is called finding the general solutions for these equations.

We still use the Unit Circle to do this, but we have to think about adding and subtracting multiples of 2π for the sin, cos, csc, and sec functions (since 2π is the period for them), and π for the tan and cot functions (since π is the period for them). We can do this by adding 2πk or πk where k is any integer (positive or negative).

Also note that a lot of times, when we get the solutions for tan, they are 180° or π radians apart, so one set of solutions will the same as the other, and we can collapse into one solution and add πk. This will sometimes happen if trig functions are squared in the problems also, since we’ll getting plusses and minuses.

Here are examples; find the general solution, or all real solutions) for the following equations.  Note that k represents all integers  \(\left( k\in \mathbb{Z} \right)\). Note also that I’m using “fancy” notation; you may not be required to do this.

Also note that sometimes you’ll solve for x or another variable, sometimes for θ, depending on your book or teacher.


Note that we need to be careful about domain restrictions with our answers.  For tan, cot, csc, and sec, we have asymptotes, and if our answer happens to fall on an asymptote, we have to eliminate it.

Solving Trig Problems:  General Solutions

 \(\displaystyle 6\cos \theta =3\sqrt{3}\)


\(\displaystyle \begin{array}{c}\cos \theta =\frac{{\sqrt{3}}}{2}\\\\\left\{ {\theta |\theta =\frac{\pi }{6}+2\pi k,\,\,\theta =\frac{{11\pi }}{6}+2\pi k} \right\}\end{array}\)

\(\displaystyle \begin{array}{c}\color{#800000}{{\sin x+3=4\sin x}}\\\\3\sin x=3\\\sin x=1\\\,\\\left\{ {x|x=\frac{\pi }{2}+\,\,2\pi k} \right\}\end{array}\) \(\displaystyle \begin{array}{c}\color{#800000}{{\sqrt{3}\csc x=-2}}\\\text{csc}\,x=-\frac{2}{{\sqrt{3}}}\,\,\,\,\,\,\,\,\,\left( {\sin x=-\frac{{\sqrt{3}}}{2}} \right)\\\\\left\{ {x|x=\frac{{4\pi }}{3}+2\pi k,\,x=\frac{{5\pi }}{3}+2\pi k} \right\}\end{array}\)
\(\displaystyle 2\sec \left( {x-\frac{\pi }{4}} \right)=4\)


\(\displaystyle \begin{array}{c}\sec \left( {x-\frac{\pi }{4}} \right)=2\text{ }\left( {\cos \left( {x-\frac{\pi }{4}} \right)=\frac{1}{2}} \right)\\\\x-\frac{\pi }{4}=\frac{\pi }{3}+2\pi k;\,\,x-\frac{\pi }{4}=\frac{{5\pi }}{3}+2\pi k\\\left\{ {x|x=\frac{{7\pi }}{{12}}+2\pi k,\,\,x=\frac{{23\pi }}{{12}}+2\pi k} \right\}\end{array}\)

\(\displaystyle \begin{array}{c}\color{#800000}{\begin{array}{c}4\sec \theta +10=-\sec \theta \\\text{(degrees)}\end{array}}\\\\5\sec \theta =-10\\\sec \theta =-2\text{ }\left( {\cos \theta =-\frac{1}{2}} \right)\\\\\left\{ \begin{array}{l}\theta |\theta =120{}^\circ +360{}^\circ k,\,\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,240{}^\circ +360{}^\circ k\end{array} \right\}\end{array}\)

\(\displaystyle \begin{array}{c}\color{#800000}{{2\sin x=4\sin x}}\\\\2\sin x=0\\\sin x=0\\\\\{x|x=2\pi k,\,\,\,x=\pi +2\pi k\}\end{array}\)

Note that (by looking at Unit Circle) this can be simplified to

\(\displaystyle \{x|x=\pi k\}\)

\(\displaystyle \text{co}{{\text{s}}^{2}}\theta =\frac{1}{2}\)


\(\displaystyle \begin{array}{c}\sqrt{{\text{co}{{\text{s}}^{2}}\theta }}=\sqrt{{\frac{1}{2}}}\\\cos \theta =\,\,\pm \frac{1}{{\sqrt{2}}}=\,\,\pm \frac{{\sqrt{2}}}{2}\\\\\{\theta |\theta =\frac{\pi }{4}+2\pi k,\,\,\,\theta =\frac{{3\pi }}{4}+2\pi k,\\\,\,\,\,\,\,\,\,\,\theta =\frac{{5\pi }}{4}+2\pi k,\,\,\theta =\frac{{7\pi }}{4}+2\pi k\}\end{array}\)

Note that (by looking at Unit Circle) this can be simplified to

\(\displaystyle \{\theta |\theta =\frac{\pi }{4}+\pi k,\,\,\,\theta =\frac{{3\pi }}{4}+\pi k\}\)

\(\displaystyle \tan \left( {x+\frac{\pi }{2}} \right)=\sqrt{3}\)


\(\displaystyle \begin{array}{c}x+\frac{\pi }{2}=\frac{\pi }{3}\,\,\,\,\,\,\,\,\,\,\,\,x+\frac{\pi }{2}=\frac{{4\pi }}{3}\\\,\,\,x=-\frac{\pi }{6}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x=\frac{{5\pi }}{6}\end{array}\)

Note that (by looking at Unit Circle) this can be simplified to

\(\displaystyle \{x|x=\frac{{5\pi }}{6}+\pi k\}\)


(Remember that you add \(\pi k\) instead of \(2\pi k\) for tan and cot).

\(\displaystyle \begin{array}{c}\color{#800000}{{3{{{\cot }}^{2}}\theta =1}}\\\text{co}{{\text{t}}^{2}}\theta =\frac{1}{3}\\\sqrt{{{{{\cot }}^{2}}\theta }}=\sqrt{{\frac{1}{3}}}\\\cot \theta =\,\,\pm \frac{1}{{\sqrt{3}}}\\\\\{\theta |\theta =\frac{\pi }{3}+\pi k,\,\,\,\theta =\frac{{2\pi }}{3}+\pi k,\\\,\,\,\,\,\,\,\,\,\theta =\frac{{4\pi }}{3}+\pi k,\,\,\theta =\frac{{5\pi }}{3}+\pi k\}\end{array}\)

Note that (by looking at Unit Circle) this can be simplified to

\(\displaystyle \{\theta |\theta =\frac{\pi }{3}+\pi k,\,\,\,\theta =\frac{{2\pi }}{3}+\pi k\}\)

\(\displaystyle \frac{{\cot \theta }}{4}=\sin \theta \cos \theta \) 


\(\displaystyle \begin{array}{c}\frac{{\cos \theta }}{{4\sin \theta }}=\sin \theta \cos \theta \\\cos \theta =4{{\sin }^{2}}\theta \cos \theta \\4{{\sin }^{2}}\theta \cos \theta -\cos \theta =0\\\cos \theta \left( {4{{{\sin }}^{2}}-1} \right)=0\\\cos \theta \left( {2\sin -1} \right)\left( {2\sin \theta +1} \right)=0\\\cos \theta =0\text{ }\,\,\text{ }\,\,\sin \theta =\pm \frac{1}{2}\,\\\\\{\theta |\theta =\frac{\pi }{2}+\pi k,\,\,\,\theta =\frac{\pi }{6}+\pi k,\,\,\,\theta =\frac{{5\pi }}{6}+\pi k\,\}\end{array}\)

Watch domain restriction:

\(\displaystyle \frac{{1-\cos x}}{{\sin x}}=\frac{{\sin x}}{{1+\cos x}}\)


\(\begin{array}{c}\left( {1-\cos x} \right)\left( {1+\cos x} \right)={{\sin }^{2}}x\\1-{{\cos }^{2}}x={{\sin }^{2}}x\\{{\sin }^{2}}x={{\sin }^{2}}x;\,\,\,\,\mathbb{R}\end{array}\)


We get all real numbers since this is an identity. But we still need to check the domain restrictions (denominator can’t be 0):


\(\begin{array}{c}\sin x\ne 0,\,\,\,1+\cos x\ne 0\\\sin x\ne 0,\,\,\,\cos x\ne -1\\\text{Solution: }\left\{ {x|\,\,x\ne \pi +2\pi k} \right\}\end{array}\)

Note that you can check these in a graphing calculator (radian mode) by putting the left-hand side of the equation into \({{Y}_{1}}\) and the right-hand side into \({{Y}_{2}}\) and get the intersection. You won’t get the exact answers, but you can still compare to the exact answers you got above.

Solving Trigonometric Equations with Multiple Angles

We have to be careful when solving trig equations with multiple angles, meaning there is a coefficient before the x or θ (variable).  This is because we could have fewer or more solutions in the Unit Circle, and thus for all real solutions when we add the 2πk or πk.

So when we solve these types of trig problems, we always want to solve for the General Solution first (even if we’re asked to get the solutions between 0 and 2π) and then go back and see how many solutions are on the Unit Circle (between 0 and 2π ).

When solving trig equations with multiple angles between 0 and 2π, we’ll typically get fewer solutions if the coefficient of the variable is less than 1, or more solutions if the coefficient of the variable is greater than 1.   As an example, we typically get 2 solutions for \(\cos \left( \theta \right)\)  between 0 and 2π, so for  \(\cos \left( 3\theta \right)\), we’ll get 2 times 3, or 6 solutions.  As another example, for  \(\cos \left( \frac{\theta }{2} \right)\), we’ll only get one solution instead of the normal two.

Let’s do some problems, finding the general solutions first, and then finding the solutions in the 0 to 2π interval.

Note that when we multiply or divide to get the variable by itself, we have to do the same with the “+2πk” or “+πk”.

Again, watch for domain restrictions; answers that happen to fall on an asymptote for tan, cot, sec, or csc.

Solving Trig Problems with Multiple Angles General Solutions

Now let’s solve the same multiple angle problems, but get solutions between 0 and 2π.


Note again that when we solve these types of trig problems, we always want to solve for the General Solution first, and then go back and see how many solutions are on the Unit Circle (between 0 and 2π ).

Solving Trig Problems with Multiple Angles 0 to 2pi

Factoring to Solve Trigonometric Equations

Note that sometimes we have to factor the equations to get the solutions, typically if they are trig quadratic equations.   Then we set all factors to 0 to solve, making sure we test the answers to see if they work.  We learned how to factor Quadratic Equations in the Solving Quadratics by Factoring and Completing the Square section.

Note that when we factor trig equations to find solutions, like we do with “regular” equations, we never just divide a factor out from each side.  In doing this, we are probably “throwing away” valid solutions to the equation.

Here are some examples, both solving on the interval 0 to 2π and over the reals; note one of the problems is using degrees instead of radians.

You will notice in the last problem that the answer in the left column ($\theta = \frac{{\pi k}}{2}$) has to be “thrown out”, because of our domain restriction for cot (it falls on an asymptote). Thus this answer is an extraneous solution:

Factoring To Solve Trig Problems

Solving Trigonometric Equations on the Calculator

We can use the graphing calculator to check these answers and also to solve trig equations that do not involve special angles.

We can put the left-hand part of the equation in  \({{Y}_{1}}\)  and the right-hand part of the equation in  \({{Y}_{2}}\)  and solve for the intersection(s) between 0 and 2π. For solving over the real (general solutions), you can put the period of the trig function and then add the appropriate factors of π, 2π, or whatever the period of the function is.

Remember for the reciprocal functions, take the reciprocal of what’s on the right hand side, and use the regular trig functions.


Always do these problems in radian mode. Also remember to use the TRACE button to move the cursor closer to the points of intersection. You may have to play around with the \(y\) portion of the window, but, again, use Xmin = 0, and Xmax = 2π when getting solutions in that interval, and use Xmin = 0 and Xmax = the period (such as  \(\frac{2\pi }{5}\)  when you have  \(\sin \left( {5x} \right)\), for example) for general solutions.

Also remember that  \({{\left( \cos \theta \right)}^{2}}\)  is written as \({{\cos }^{2}}\theta \), and we can put it in the graphing calculator as \(\boldsymbol{\cos {{\left( x \right)}^{2}}}\) or \(\boldsymbol {{{\left( {\cos \left( x \right)} \right)}^{2}}}\).

Here are some examples:

Solving Trig Problems with a Calculator

Solving Trig Systems of Equations

We learned how to solve systems of more complicated equations here in the Systems of non-Linear Equations section.

Again, we can use either Substitution or Elimination, depending on what’s easier.  Once we get the initial solution(s), we’ll need to plug in to get the other variable.

Here are some examples of Solving Systems with Trig Equations; solve over the reals:

Solving Systems of Equations with Trig

Solving Trig Inequalities

Sometimes you might be asked to solve a Trig Inequality.  There are links to many other types of Inequalities here (we saw one of these above).

We can either solve these inequalities graphically or algebraically; let’s try one of each.  Note that you can also solve these on your graphing calculator, using the Intersect feature, and then see where the inequalities “work”:

Trig Inequalities

Practice these problems, and practice, practice, practice!

Hit Submit (the arrow to the right of the problem) to solve this problem. You can also type in more problems, or click on the 3 dots in the upper right hand corner to drill down for example problems.

For Practice: Use the Mathway widget below to try a Trig Solving problem. Click on Submit (the blue arrow to the right of the problem) and click on Solve for x to see the answer.

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On to Trigonometric Identities – you’re ready!