This section covers:
 Solving Trigonometric Equations Using the Unit Circle
 Solving Trigonometric Equations – General Solutions
 Solving Trigonometric Equations with Multiple Angles
 Factoring to Solve Trigonometric Equations
 Solving Trigonometric Equations Using a Calculator
 Solving Trig Systems of Equations
 Solving Trig Inequalities
 More Practice
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 \(\theta \) 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 \(\theta \) 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 \(\left[ 0,2\pi \right)\) interval. We will learn how to do this here.
Note that sometimes you may have to solve using degrees \(\left[ {0,{{{360}}^{o}}} \right)\) 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: \(\boldsymbol {\left[ {0,2\pi } \right)}\) or \(\boldsymbol{\left[ {0,360{}^\circ } \right)}\)  
\(\displaystyle \sin \theta =1\) \(\displaystyle \theta =\left\{ {\frac{\pi }{2}} \right\}\)
\(\displaystyle \sin \theta =0\) \(\theta =\left\{ {0,\pi } \right\}\)
\(\displaystyle \cos \theta =0\) \(\displaystyle \theta =\left\{ {\frac{\pi }{2},\frac{{3\pi }}{2}} \right\}\) 
\(\tan \theta +1=0\)
\(\displaystyle \tan \theta =1\) \(\displaystyle \theta =\left\{ {\frac{{3\pi }}{4},\frac{{7\pi }}{4}} \right\}\)
\(\tan \theta 1=0\) \(\displaystyle \tan \theta =1\) \(\displaystyle \theta =\left\{ {\frac{\pi }{4},\frac{{5\pi }}{4}} \right\}\) 
\(\displaystyle \begin{array}{c}3\sqrt{2}\cos \theta +3=0\\\text{(degrees)}\end{array}\)
\(\displaystyle \begin{align}3\sqrt{2}\cos \theta &=3\\\cos \theta &=\frac{3}{{3\sqrt{2}}}\\\cos \theta &=\frac{1}{{\sqrt{2}}}=\frac{{\sqrt{2}}}{2}\\\theta &=\left\{ {135{}^\circ ,225{}^\circ } \right\}\end{align}\) 
\(2\sin \theta +4=3\)
\(\displaystyle \begin{array}{c}2\sin \theta =1\\\sin \theta =\frac{1}{2}\end{array}\) \(\displaystyle \theta =\left\{ {\frac{{7\pi }}{6},\frac{{11\pi }}{6}} \right\}\) 
\(\displaystyle 4\sec \theta +7=1\)
\(\displaystyle \begin{array}{c}4\sec \theta =8\\\sec \theta =2\,\,\,\,\,\left( {\cos \theta =\frac{1}{2}} \right)\end{array}\) \(\displaystyle \theta =\left\{ {\frac{{2\pi }}{3},\frac{{4\pi }}{3}} \right\}\) 
\(\displaystyle 2{{\cos }^{2}}\theta =1\)
\(\displaystyle \begin{array}{c}{{\cos }^{2}}\theta =\frac{1}{2}\\\sqrt{{{{{\cos }}^{2}}\theta }}=\sqrt{{\frac{1}{2}}}\\\cos \theta =\pm \frac{1}{{\sqrt{2}}}=\pm \frac{{\sqrt{2}}}{2}\end{array}\) \(\displaystyle \theta =\left\{ {\frac{\pi }{4},\frac{{3\pi }}{4},\frac{{5\pi }}{4},\frac{{7\pi }}{4}} \right\}\) 
\(\displaystyle \sin \left( {\theta +\frac{\pi }{{18}}} \right)=0\)
\(\displaystyle \begin{align}\theta +\frac{\pi }{{18}}=0\,\,\,\,\,\,&\,\,\,\,\theta +\frac{\pi }{{18}}=\pi \\\,\,\,\theta =\frac{\pi }{{18}}\,\,\,\,\,\,\,&\,\,\,\,\,\,\,\theta =\frac{{17\pi }}{{18}}\end{align}\) Since \(\displaystyle \frac{\pi }{{18}}\) isn’t in \(\left[ {0,2\pi } \right)\), \(\displaystyle \theta =\left\{ {\frac{\pi }{{18}}+2\pi ,\frac{{17\pi }}{{18}}} \right\}\) \(\displaystyle \theta =\left\{ {\frac{{17\pi }}{{18}},\frac{{35\pi }}{{18}}} \right\}\) 
\(\displaystyle \sin \left( {\frac{\theta }{3}+\frac{\pi }{3}} \right)=\cos \left( {\frac{\theta }{3}+\frac{\pi }{3}} \right)\)
\(\displaystyle \tan \left( {\frac{\theta }{3}+\frac{\pi }{3}} \right)=1\) \(\displaystyle \begin{align}\,\frac{\theta }{3}+\frac{\pi }{3}=\frac{\pi }{4}\,\,\,\,\,\,&\,\,\,\,\,\,\frac{\theta }{3}+\frac{\pi }{3}=\frac{{5\pi }}{4}\\\,\,\,\frac{\theta }{3}=\frac{\pi }{{12}}\,\,\,\,\,\,\,\,&\,\,\,\,\,\,\,\,\frac{\theta }{3}=\frac{{11\pi }}{{12}}\\\,\,\,\theta =\frac{\pi }{4}\,\,\,\,\,\,\,\,\,&\,\,\,\,\,\,\,\,\,\theta =\frac{{33\pi }}{{12}}\end{align}\) Neither are in \(\left[ {0,2\pi } \right)\), and adding \(2\pi \) to \(\displaystyle \frac{\pi }{4}\) doesn’t work (extraneous); no solution, or \(\emptyset\). 
Here’s one where we have an inequality:
For what values of \(\theta \) is \(2\cos \theta <\sqrt{3}\) on the \(\left[ {0,2\pi } \right)\) interval?

\(\displaystyle 2\cos \theta <\sqrt{3}\) \(\displaystyle \cos \theta <\frac{{\sqrt{3}}}{2}\) This is a little tricky; if we look on the Unit Circle, we want the cos values to be less than where \(\displaystyle \cos \theta =\frac{{\sqrt{3}}}{2}\). Since when \(\displaystyle \cos \theta =\frac{{\sqrt{3}}}{2},\,\,\theta =\left\{ {\frac{{5\pi }}{6},\frac{{7\pi }}{6}} \right\}\), the \(x\) will be less (or to the left of) of those 2 values, so we can see that \(\displaystyle \frac{{5\pi }}{6}<\theta <\frac{{7\pi }}{6}\text{ or }\left( {\frac{{5\pi }}{6},\frac{{7\pi }}{6}} \right)\).
You can also plug in values for cos on the Unit Circle to see which are less than \(\displaystyle \frac{{\sqrt{3}}}{2}\), or use a Graphing Calculator to graph the two sides of the inequality.

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 \(2\pi \) for the sin, cos, csc, and sec functions (since \(2\pi \) is the period for them), and \(\pi \) for the tan and cot functions (since \(\pi \) is the period for them). We can do this by adding \(2\pi k\) or \(\pi k\) 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 \(\pi \) radians apart, so one set of solutions will the same as the other, and we can collapse into one solution and add \(\pi k\). 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 \(\theta \), 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 

\(\displaystyle 6\cos \theta =3\sqrt{3}\)
\(\displaystyle \cos \theta =\frac{{\sqrt{3}}}{2}\) \(\displaystyle \left\{ {\theta \theta =\frac{\pi }{6}+2\pi k,\,\,\theta =\frac{{11\pi }}{6}+2\pi k} \right\}\) 
\(\displaystyle \sin x+3=4\sin x\)
\(\displaystyle \begin{array}{c}3\sin x=3\\\sin x=1\end{array}\) \(\displaystyle \left\{ {xx=\frac{\pi }{2}+\,\,2\pi k} \right\}\) 
\(\displaystyle \sqrt{3}\csc x=2\)
\(\displaystyle \text{csc}\,x=\frac{2}{{\sqrt{3}}}\,\,\,\,\,\,\,\,\,\,\left( {\sin x=\frac{{\sqrt{3}}}{2}} \right)\) \(\displaystyle \left\{ {xx=\frac{{4\pi }}{3}+2\pi k,\,\,x=\frac{{5\pi }}{3}+2\pi k} \right\}\) 

\(\displaystyle 2\sec \left( {x\frac{\pi }{4}} \right)=4\)
\(\displaystyle \sec \left( {x\frac{\pi }{4}} \right)=2\text{ }\left( {\cos \left( {x\frac{\pi }{4}} \right)=\frac{1}{2}} \right)\) \(\displaystyle x\frac{\pi }{4}=\frac{\pi }{3}+2\pi k;\,\,\,x\frac{\pi }{4}=\frac{{5\pi }}{3}+2\pi k\) \(\displaystyle \left\{ {xx=\frac{{7\pi }}{{12}}+2\pi k,\,\,x=\frac{{23\pi }}{{12}}+2\pi k} \right\}\) 
\(\displaystyle \begin{array}{c}4\sec \theta +10=\sec \theta \\\text{(degrees)}\end{array}\)
\(\displaystyle \begin{array}{c}5\sec \theta =10\\\sec \theta =2\text{ }\left( {\cos \theta =\frac{1}{2}} \right)\\\left\{ \begin{array}{l}\theta \theta =120{}^\circ +360{}^\circ k,\,\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,240{}^\circ +360{}^\circ k\end{array} \right\}\end{array}\) 
\(\displaystyle 2\sin x=4\sin x\)
\(\displaystyle \begin{array}{c}2\sin x=0\\\sin x=0\\\{xx=2\pi k,\,\,\,x=\pi +2\pi k\}\end{array}\) Note that (by looking at Unit Circle) this can be simplified to \(\displaystyle \{xx=\pi k\}\) 

\(\displaystyle \text{co}{{\text{s}}^{2}}\theta =\frac{1}{2}\)
\(\displaystyle \begin{align}\sqrt{{\text{co}{{\text{s}}^{2}}\theta }}&=\sqrt{{\frac{1}{2}}}\\\cos \theta &=\pm \frac{1}{{\sqrt{2}}}=\pm \frac{{\sqrt{2}}}{2}\\\{\theta \theta =\frac{\pi }{4}+2\pi k,\,&\,\theta =\frac{{3\pi }}{4}+2\pi k,\\\,\,\theta =\frac{{5\pi }}{4}+2\pi k,\,&\,\theta =\frac{{7\pi }}{4}+2\pi k\}\end{align}\) Note that (by looking at Unit Circle) this can be simplified to \(\displaystyle \{\theta \theta =\frac{\pi }{4}+\pi k,\,\,\,\theta =\frac{{3\pi }}{4}+\pi k\}\) 
\(\displaystyle \tan \left( {x+\frac{\pi }{2}} \right)=\sqrt{3}\)
\(\displaystyle \begin{align}x+\frac{\pi }{2}=\frac{\pi }{3}\,\,\,\,\,&\,\,\,\,\,x+\frac{\pi }{2}=\frac{{4\pi }}{3}\\\,\,\,x=\frac{\pi }{6}\,\,\,\,\,\,\,\,&\,\,\,\,\,\,\,x=\frac{{5\pi }}{6}\end{align}\) Note that (by looking at Unit Circle) this can be simplified to \(\displaystyle \{xx=\frac{{5\pi }}{6}+\pi k\}\)
(Remember that you add \(\pi k\) instead of \(2\pi k\) for tan and cot). 
\(\displaystyle 3{{\cot }^{2}}\theta =1\)
\(\displaystyle \begin{align}\text{co}{{\text{t}}^{2}}\theta &=\frac{1}{3}\\\sqrt{{{{{\cot }}^{2}}\theta }}&=\sqrt{{\frac{1}{3}}}\\\cot \theta &=\pm \frac{1}{{\sqrt{3}}}\\\{\theta \theta =\frac{\pi }{3}+\pi k,\,\,&\,\theta =\frac{{2\pi }}{3}+\pi k,\\\,\,\,\theta =\frac{{4\pi }}{3}+\pi k,\,\,&\,\theta =\frac{{5\pi }}{3}+\pi k\}\end{align}\) Note that (by looking at Unit Circle) this can be simplified to \(\displaystyle \{\theta \theta =\frac{\pi }{3}+\pi k,\,\,\,\theta =\frac{{2\pi }}{3}+\pi k\}\) 

\(\displaystyle \frac{{\cot \theta }}{4}=\sin \theta \cos \theta \)
\(\displaystyle \begin{array}{c}\frac{{\cos \theta }}{{4\sin \theta }}=\sin \theta \cos \theta \\\cos \theta =4{{\sin }^{2}}\theta \cos \theta \\4{{\sin }^{2}}\theta \cos \theta \cos \theta =0\\\cos \theta \left( {4{{{\sin }}^{2}}1} \right)=0\\\cos \theta \left( {2\sin 1} \right)\left( {2\sin \theta +1} \right)=0\\\cos \theta =0\text{ }\,\,\text{ }\,\,\sin \theta =\pm \frac{1}{2}\end{array}\) \(\displaystyle \{\theta \theta =\frac{\pi }{2}+\pi k,\,\,\,\theta =\frac{\pi }{6}+\pi k,\,\,\,\theta =\frac{{5\pi }}{6}+\pi k\,\}\) 
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 lefthand side of the equation into \({{Y}_{1}}\) and the righthand 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 \(\theta \) (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 \(2\pi k\) or \(\pi k\).
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 \(2\pi k\)) and then go back and see how many solutions are on the Unit Circle (between 0 and \(2\pi k\) ).
When solving trig equations with multiple angles between 0 and \(2\pi \), 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 \(2\pi \), so for \(\cos \left( 3\theta \right)\), we’ll get 2 times 3, or 6 solutions. As another example, for \(\displaystyle \cos \left( \frac{\theta }{2} \right)\), 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 \(2\pi \) interval.
Note that when we multiply or divide to get the variable by itself, we have to do the same with the “\(+2\pi k\)” or “\(+\pi k\)”.
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  
\(\displaystyle \tan \left( {3\theta } \right)=\sqrt{3}\)
\(\displaystyle 3\theta =\frac{\pi }{3}+\pi k\,\,\,\,\,\,\,\,3\theta =\frac{{4\pi }}{3}+\pi k\)
Looking at Unit Circle, this can be simplified to: \(\displaystyle \begin{align}3\theta =\frac{\pi }{3}+\pi k\,\,\,\,\\\left\{ {\theta \theta =\frac{\pi }{9}+\frac{\pi }{3}k} \right\}\end{align}\) 
\(\displaystyle 2\cos \left( {3x} \right)+\sqrt{3}=0\)
\(\displaystyle \begin{align}\cos \left( {3x} \right)&=\frac{{\sqrt{3}}}{2}\\\,\,\,3x=\frac{{5\pi }}{6}+2\pi k\,\,\,\,\,\,&\,\,\,\,\,\,3x=\frac{{7\pi }}{6}+2\pi k\end{align}\) \(\displaystyle \left\{ {xx=\frac{{5\pi }}{{18}}+\frac{2}{3}\pi k,\,\,x=\frac{{7\pi }}{{18}}+\frac{2}{3}\pi k\,\,} \right\}\) 
\(\displaystyle \sqrt{2}\sec \left( {\frac{x}{6}} \right)2=0\)
(degrees)
\(\displaystyle \begin{align}\sec \left( {\frac{x}{6}} \right)=\frac{2}{{\sqrt{2}}}\,\,\,\,&\,\,\,\,\left( {\cos \left( {\frac{x}{6}} \right)=\frac{{\sqrt{2}}}{2}} \right)\\\,\,\frac{x}{6}=45{}^\circ +360{}^\circ k\,\,\,\,\,\,\,&\,\,\,\,\,\,\,\frac{x}{6}=315{}^\circ +360{}^\circ k\\\,\,\,x=270{}^\circ +2160{}^\circ k\,\,\,\,\,\,\,&\,\,\,\,\,\,\,x=1890{}^\circ +2160{}^\circ k\end{align}\) \(\displaystyle \left\{ {xx=270{}^\circ +2160{}^\circ k,\,\,\,\,x=1890{}^\circ +2160{}^\circ k} \right\}\) By doing some subtraction from \(\displaystyle 1890{}^\circ \), this can be simplified to: \(\displaystyle \left\{ {xx=\pm 270{}^\circ +2160{}^\circ k} \right\}\) 
\(\displaystyle 5{{\cos }^{2}}\left( {\frac{\theta }{3}} \right)=5\)
\(\displaystyle \begin{align}\sqrt{{{{{\cos }}^{2}}\left( {\frac{\theta }{3}} \right)}}&=\sqrt{1}\\\cos \left( {\frac{\theta }{3}} \right)&=\pm 1\\\,\,\,\frac{\theta }{3}=0+2\pi k\,\,\,\,\,\,&\,\,\,\,\frac{\theta }{3}=\pi +2\pi k\end{align}\)
Looking at Unit Circle, this can be simplified to: \(\displaystyle \begin{align}\frac{\theta }{3}=\pi k\\\left\{ {\theta \theta =3\pi k} \right\}\end{align}\) 
\(\displaystyle 2{{\sin }^{2}}\left( {2x} \right)=1\)
\(\displaystyle \begin{align}{{\sin }^{2}}\left( {2x} \right)&=\frac{1}{2}\\\sqrt{{{{{\sin }}^{2}}\left( {2x} \right)}}&=\sqrt{{\frac{1}{2}}}\\\sin \left( {2x} \right)&=\,\,\pm \frac{1}{{\sqrt{2}}}=\,\,\pm \frac{{\sqrt{2}}}{2}\\\,\,\,2x=\frac{\pi }{4}+2\pi k\,\,\,\,\,\,&\,\,\,\,\,\,2x=\frac{{3\pi }}{4}+2\pi k\\\,\,\,2x=\frac{{5\pi }}{4}+2\pi k\,\,\,\,\,&\,\,\,\,\,2x=\frac{{7\pi }}{4}+2\pi k\end{align}\)
Looking at Unit Circle, this can be simplified to: \(\displaystyle \,2x=\frac{\pi }{4}+\pi k\,\,\,\,\,\,\,\,\,\,\,\,2x=\frac{{3\pi }}{4}+\pi k\) \(\displaystyle \left\{ {xx=\frac{\pi }{8}+\frac{\pi }{2}k,\,\,\,x=\frac{{3\pi }}{8}+\frac{\pi }{2}k} \right\}\) 
\(\displaystyle \tan \left( {\frac{\theta }{2}\frac{\pi }{3}} \right)=1\)
\(\displaystyle \frac{\theta }{2}\frac{\pi }{3}=\frac{{3\pi }}{4}+\pi k\,\,\,\,\,\,\,\,\,\frac{\theta }{2}\frac{\pi }{3}=\frac{{7\pi }}{4}+\pi k\)
Looking at Unit Circle, this can be simplified to: \(\displaystyle \begin{align}\,\frac{\theta }{2}\frac{\pi }{3}&=\frac{{3\pi }}{4}+\pi k\\\frac{\theta }{2}&=\left( {\frac{{3\pi }}{4}+\frac{\pi }{3}} \right)+\pi k\\\frac{\theta }{2}&=\frac{{13\pi }}{{12}}+\pi k\end{align}\) \(\displaystyle \left\{ {\theta \theta =\frac{{13\pi }}{6}+2\pi k} \right\}\,\,\,\text{or}\,\,\left\{ {\theta \theta =\frac{\pi }{6}+2\pi k} \right\}\) 
Now let’s solve the same multiple angle problems, but get solutions between 0 and \(2\pi k\) or \(360{}^\circ \).
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 \(2\pi k\)).
Solving Trig Problems with Multiple Angles: \(\boldsymbol {\left[ {0,2\pi } \right)}\) or \(\boldsymbol{\left[ {0,360{}^\circ } \right)}\)  
\(\displaystyle \tan \left( {3\theta } \right)=\sqrt{3}\)
\(\displaystyle 3\theta =\frac{\pi }{3}+\pi k\,\,\,\,\,\,\,\,3\theta =\frac{{4\pi }}{3}+\pi k\)
Looking at Unit Circle, this can be simplified to: \(\displaystyle \begin{align}3\theta &=\frac{\pi }{3}+\pi k\\\,\theta &=\frac{\pi }{9}+\frac{\pi }{3}k=\frac{\pi }{9}+\frac{{3\pi }}{9}k\end{align}\)
Looking at the Unit Circle, we can get solutions between 0 and \(2\pi \) by adding \(\displaystyle \frac{{3\pi }}{9}\), staying under \(2\pi \): \(\displaystyle \theta =\left\{ {\frac{\pi }{9},\frac{{4\pi }}{9},\frac{{7\pi }}{9},\frac{{10\pi }}{9},\frac{{13\pi }}{9},\frac{{16\pi }}{9}} \right\}\) 
\(\displaystyle 2\cos \left( {3x} \right)+\sqrt{3}=0\)
\(\displaystyle \begin{align}\cos \left( {3x} \right)&=\frac{{\sqrt{3}}}{2}\\\,\,\,3x=\frac{{5\pi }}{6}+2\pi k\,\,\,\,&\,\,\,\,\,3x=\frac{{7\pi }}{6}+2\pi k\\\,\,\,x=\frac{{5\pi }}{{18}}+\frac{{12}}{{18}}\pi k\,\,\,\,\,&\,\,\,\,\,x=\frac{{7\pi }}{{18}}+\frac{{12}}{{18}}\pi k\end{align}\)
Looking at the Unit Circle, we can get solutions between 0 and \(2\pi \) by adding \(\displaystyle \frac{{12\pi }}{{18}}\), staying under \(2\pi \): \(\displaystyle x=\left\{ {\frac{{5\pi }}{{18}},\frac{{17\pi }}{{18}},\frac{{29\pi }}{{18}},\frac{{7\pi }}{{18}},\frac{{19\pi }}{{18}},\frac{{31\pi }}{{18}}} \right\}\) 
\(\displaystyle \sqrt{2}\sec \left( {\frac{x}{6}} \right)2=0\)
\(\displaystyle \begin{align}\sec \left( {\frac{x}{6}} \right)=\frac{2}{{\sqrt{2}}}\,\,\,\,&\,\,\,\left( {\cos \left( {\frac{x}{6}} \right)=\frac{{\sqrt{2}}}{2}} \right)\\\,\,\frac{x}{6}=\frac{\pi }{4}+2\pi k\,\,\,\,\,\,&\,\,\,\,\,\frac{x}{6}=\frac{{7\pi }}{4}+2\pi k\\\,\,\,x=\frac{{6\pi }}{4}+12\pi k\,\,\,\,\,\,&\,\,\,\,\,\,x=\frac{{42\pi }}{4}+12\pi k\end{align}\)
Looking at the Unit Circle, the only solution that will work is \(\displaystyle \frac{{6\pi }}{4}=\frac{{3\pi }}{2}\). \(\displaystyle x=\left\{ {\frac{{3\pi }}{2}} \right\}\) 
\(\displaystyle 5{{\cos }^{2}}\left( {\frac{\theta }{3}} \right)=5\) (degrees)
\(\displaystyle \begin{align}\sqrt{{{{{\cos }}^{2}}\left( {\frac{\theta }{3}} \right)}}&=\sqrt{1}\\cos\left( {\frac{\theta }{3}} \right)&=\pm 1\\\,\,\,\frac{\theta }{3}=0+360{}^\circ k\,\,\,\,\,&\,\,\,\,\,\frac{\theta }{3}=180{}^\circ +360{}^\circ k\end{align}\)
Looking at the Unit Circle, this can be simplified to: \(\displaystyle \begin{align}\frac{\theta }{3}&=180{}^\circ k\\\theta &=540{}^\circ k\end{align}\)
Looking at the Unit Circle, the only solution that will work is 0. \(\displaystyle \theta =\left\{ 0 \right\}\) 
\(\displaystyle 2{{\sin }^{2}}\left( {2x} \right)=1\)
\(\displaystyle \begin{align}\sqrt{{{{{\sin }}^{2}}\left( {2x} \right)}}&=\sqrt{{\frac{1}{2}}}\\\sin \left( {2x} \right)&=\pm \frac{1}{{\sqrt{2}}}=\pm \frac{{\sqrt{2}}}{2}\\\,2x=\frac{\pi }{4}+2\pi k\,\,\,\,&\,\,\,\,2x=\frac{{3\pi }}{4}+2\pi k\\\,\,\,2x=\frac{{5\pi }}{4}+2\pi k\,\,\,\,&\,\,\,\,2x=\frac{{7\pi }}{4}+2\pi k\end{align}\)
Looking at the Unit Circle, this can be simplified to: \(\displaystyle \begin{align}2x=\frac{\pi }{4}+\pi k\,\,\,\,\,\,&\,\,\,\,2x=\frac{{3\pi }}{4}+\pi k\\\,\,x=\frac{\pi }{8}+\frac{\pi }{2}k,\,\,&\,\,x=\frac{{3\pi }}{8}+\frac{\pi }{2}k\end{align}\)
Looking at the Unit Circle, we can get solutions between 0 and \(2\pi \) by adding \(\displaystyle \frac{{4\pi }}{8}\), staying under \(2\pi \): \(\displaystyle x=\left\{ \begin{align}&\frac{\pi }{8},\frac{{5\pi }}{8},\frac{{9\pi }}{8},\frac{{13\pi }}{8},\\&\frac{{3\pi }}{8},\frac{{7\pi }}{8},\frac{{11\pi }}{8},\frac{{15\pi }}{8}\end{align} \right\}\) 
\(\displaystyle \tan \left( {\frac{\theta }{2}\frac{\pi }{3}} \right)=1\)
\(\displaystyle \begin{align}\frac{\theta }{2}\frac{\pi }{3}&=\frac{{3\pi }}{4}+\pi k\\\frac{\theta }{2}\frac{\pi }{3}&=\frac{{7\pi }}{4}+\pi k\end{align}\)
Looking at the Unit Circle, this can be simplified to: \(\displaystyle \begin{align}\frac{\theta }{2}\frac{\pi }{3}&=\frac{{3\pi }}{4}+\pi k\\\frac{\theta }{2}&=\left( {\frac{{3\pi }}{4}+\frac{\pi }{3}} \right)+\,\,\pi k\\\frac{\theta }{2}&=\frac{{13\pi }}{{12}}+\pi k\\\theta &=\frac{{13\pi }}{6}+2\pi k\end{align}\)
Looking at the Unit Circle, the only solution that will work is: \(\begin{align}\frac{{13\pi }}{6}2\pi =\frac{\pi }{6}\\\theta =\left\{ {\frac{\pi }{6}} \right\}\end{align}\)

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 \(2\pi \) (or \(360{}^\circ \)) 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 (\(\displaystyle \theta = \frac{{\pi k}}{2}\)) 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 Equations  
\(\displaystyle \begin{array}{c}4\sin x\cos x=2\sin x\\\text{Interval }(0,2\pi )\end{array}\)
\(\displaystyle \begin{array}{c}4\sin x\cos x2\sin x=0\\2\sin x\left( {2\cos x1} \right)=0\end{array}\)
\(\displaystyle \begin{align}2\sin x=0\,\,\,\,\,\,&\,\,\,\,2\cos x1=0\\\sin x=0\,\,\,\,\,\,&\,\,\,\,\,\cos x=\frac{1}{2}\\x=0,\,\,\pi \,\,\,\,\,&\,\,\,\,x=\frac{\pi }{3},\,\frac{{5\pi }}{3}\end{align}\) \(\displaystyle x=\left\{ {0,\pi ,\frac{\pi }{3},\frac{{5\pi }}{3}} \right\}\) 
\(\displaystyle \begin{array}{c}4{{\cos }^{4}}\theta 7{{\cos }^{2}}\theta +3=0\\\text{General Solutions (in degrees)}\end{array}\)
\(\displaystyle \left( {4{{{\cos }}^{2}}\theta 3} \right)\left( {{{{\cos }}^{2}}\theta 1} \right)=0\)
\(\displaystyle \begin{align}\sqrt{{{{{\cos }}^{2}}\theta }}=\sqrt{{\frac{3}{4}}}\,\,\,\,\,\,&\,\,\,\,\,\sqrt{{{{{\cos }}^{2}}\theta }}=\sqrt{1}\\\,\cos \theta =\pm \frac{{\sqrt{3}}}{2}\,\,\,\,\,\,\,&\,\,\,\,\,\cos \theta =\pm 1\,\end{align}\) \(\displaystyle \begin{array}{c}\text{(Add }180{}^\circ k\text{ instead of }360{}^\circ k\text{ because of }\pm )\\\,\,\,\theta =30{}^\circ +180{}^\circ k\,\,\,\,\,\,\,\,\,\,\,\,\theta =180{}^\circ k\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\theta =330{}^\circ +180{}^\circ k\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\\\left\{ \begin{array}{l}\theta \theta =30{}^\circ +180{}^\circ k,\,\,\theta =330{}^\circ +180{}^\circ k,\\\,\,\,\,\theta =180{}^\circ k\end{array} \right\}\end{array}\) 
\(\displaystyle \begin{array}{c}2{{\sec }^{2}}x3\sec x=2\\\text{General Solutions}\end{array}\)
\(\displaystyle \begin{array}{c}2{{\sec }^{2}}x3\sec x2=0\\\left( {2\sec x+1} \right)\left( {\sec x2} \right)=0\end{array}\)
\(\displaystyle \begin{align}\sec x=\frac{1}{2}\,\,\,\,\,\,\,\,&\,\,\,\,\,\,\sec x=2\\\text{(no solution)}\,\,\,\,\,\,\,&\,\,\,\,\,\left( {\cos x=\frac{1}{2}} \right)\end{align}\) \(\displaystyle \left\{ {xx=\frac{\pi }{3}+2\pi k,\,x=\frac{{5\pi }}{3}+2\pi k\,} \right\}\) 
\(\displaystyle \begin{array}{c}{{\tan }^{4}}x4{{\tan }^{2}}x+3=0\\\text{General Solutions}\end{array}\)
\(\displaystyle \begin{array}{c}\left( {{{{\tan }}^{2}}x3} \right)\left( {{{{\tan }}^{2}}x1} \right)=0\\{{\tan }^{2}}x3=0\,\,\,\,\,\,\,\,\,{{\tan }^{2}}x1=0\end{array}\)
\(\displaystyle \begin{align}\sqrt{{{{{\tan }}^{2}}x}}=\sqrt{3}\,\,\,\,\,\,&\,\,\,\,\sqrt{{{{{\tan }}^{2}}x}}=\sqrt{1}\\\tan x=\pm \sqrt{3}\,\,\,\,\,\,\,&\,\,\,\,\,\tan x=\pm \sqrt{1}\\x=\frac{\pi }{3}+\pi k\,\,\,\,\,\,\,\,&\,\,\,\,\,\,x=\frac{\pi }{4}+\pi k\\x=\frac{{2\pi }}{3}+\pi k\,\,\,\,\,\,\,&\,\,\,\,\,\,x=\frac{{3\pi }}{4}+\pi k\end{align}\) \(\displaystyle \left\{ \begin{align}xx=\frac{\pi }{3}+\pi k,\,\,x=\frac{{2\pi }}{3}+\pi k,\\\,\,\,\,x=\frac{\pi }{4}+\pi k,\,\,\,x=\frac{{3\pi }}{4}+\pi k\end{align} \right\}\) 
\(\displaystyle \begin{array}{c}3{{\cot }^{2}}x+3\cot x\sqrt{3}\cot x=\sqrt{3}\\\text{Interval }(0,2\pi )\end{array}\)
\(\displaystyle \begin{array}{c}3{{\cot }^{2}}x+3\cot x\sqrt{3}\cot x\sqrt{3}=0\\3\cot x\left( {\cot x+1} \right)\sqrt{3}\left( {\cot x+1} \right)=0\\\left( {\cot x+1} \right)\left( {3\cot x\sqrt{3}} \right)=0\end{array}\)
\(\displaystyle \begin{align}\cot x=1\,\,\,\,\,\,&\,\,\,\,\cot x=\frac{{\sqrt{3}}}{3}\\\tan x=1\,\,\,\,\,&\,\,\,\,\tan x=\frac{3}{{\sqrt{3}}}=\sqrt{3}\\x=\frac{{3\pi }}{4},\,\,\frac{{7\pi }}{4}\,\,\,\,\,&\,\,\,\,\,x=\frac{\pi }{3},\,\,\frac{{4\pi }}{3}\end{align}\) \(\displaystyle x=\left\{ {\frac{{3\pi }}{4},\frac{{7\pi }}{4},\frac{\pi }{3},\frac{{4\pi }}{3}} \right\}\) 
\(\displaystyle \begin{array}{c}\sqrt{3}\sin \left( {2\theta } \right)\cot \left( {2\theta } \right)\sin \left( {2\theta } \right)=0\\\text{Interval }(0,2\pi )\end{array}\)
\(\displaystyle \sin \left( {2\theta } \right)\left( {\sqrt{3}\cot \left( {2\theta } \right)1} \right)=0\) \(\displaystyle \begin{align}\sin \left( {2\theta } \right)=0\,\,\,\,\,\,\,&\,\,\,\,\,\cot \left( {2\theta } \right)=\frac{1}{{\sqrt{3}}}\\2\theta =\pi k\,\,\,\,\,\,\,\,\,&\,\,\,\,\,\,2\theta =\frac{\pi }{3}+\pi k\\\,\,\,\,\,\,\,\,\,\,&\,\,\,\,\,\,2\theta =\frac{{4\pi }}{3}+\pi k\\\,\,\,\,\,\theta =\frac{{\pi k}}{2}\,\,\,\,\,\,&\,\,\,\,\,\,\,\theta =\frac{\pi }{6}+\frac{\pi }{2}k\\\,\,\,\,\,\,\,&\,\,\,\,\,\,\,\,\theta =\frac{{4\pi }}{6}+\frac{\pi }{2}k\end{align}\) Because of the domain restriction for cot (where its asymptotes are), and noting that \(\displaystyle \cot \left[ {2\left( {\frac{{\pi k}}{2}} \right)} \right]=\cot \left( {\pi k} \right)\) is undefined, we have to eliminate \(\displaystyle \frac{{\pi k}}{2}\). Solutions are: \(\displaystyle \theta =\left\{ {\frac{\pi }{6},\frac{{2\pi }}{3},\frac{{7\pi }}{6},\frac{{5\pi }}{3}} \right\}\) 
Solving Trigonometric Equations Using a Calculator
We already used a calculator to find inverse trig functions in a calculator here in the Inverse Trigonometric Functions section. When solving trig equations, however, it’s a little more complicated, since typically we’ll have multiple solutions.
We can do this using just a scientific calculator, or graphing the functions and finding intersections with a graphing calculator (usually easier). Don’t forget to change to the appropriate mode (radians or degrees) using DRG on a TI scientific calculator, or mode on a TI graphing calculator.
If just using a scientific calculator, here are some rules for solving trig problems in the intervals \(\left[ {0,2\pi } \right)\) or \(\left( {\infty ,\infty } \right)\) in radians (substitute 180° for if using degrees). Remember these rules, which make sense if you look at the trig functions on the Unit Circle. Remember that \(k\) is any integer, negative, 0, or positive.
 For \(\displaystyle \sin \theta =A,\,\theta ={{\sin }^{{1}}}A\,\,\,\left( {+\,2\pi k} \right)\) and also \(\displaystyle \theta =\pi {{\sin }^{{1}}}A\,\,\left( {+\,2\pi k} \right)\). (For \(\csc \theta =A\), use \(\displaystyle {{\sin }^{{1}}}\left( {\frac{1}{A}} \right)\)). For example, in the interval \(\left[ {0,2\pi } \right)\), for \(\displaystyle \sin \theta =.5,\,\,\theta ={{\sin }^{{1}}}\left( {.5\,} \right)\approx .524\,\,\left( {\frac{\pi }{6}\,\,\,\text{or}\,\,\frac{{11\pi }}{6}\,} \right)\), and also \(\displaystyle \theta =\pi {{\sin }^{{1}}}\left( {.5} \right)\,=\pi \frac{{11\pi }}{6}=\,\,\frac{{5\pi }}{6}\).
 For \(\displaystyle \cos \theta =A,\,\theta ={{\cos }^{{1}}}A\,\,\,\left( {+\,2\pi k} \right)\), and also \(\displaystyle \theta =2\pi {{\cos }^{{1}}}A\,\,\left( {+\,2\pi k} \right)\), which is the same as \(\displaystyle {{\cos }^{{1}}}A\,\,\left( {+\,2\pi k} \right)\). (For \(\sec \theta =A\), use \(\displaystyle {{\cos }^{{1}}}\left( {\frac{1}{A}} \right)\) ). For example, in the interval \(\left[ {0,2\pi } \right)\), for \(\displaystyle \cos \theta =.5,\,\,\theta ={{\cos }^{{1}}}\left( {.5} \right)\,\approx 1.047\,\,\left( {\frac{\pi }{3}} \right)\), and also \(\displaystyle \theta =2\pi {{\cos }^{{1}}}\left( {.5} \right)\,\approx 5.236\,\,\left( {\frac{{5\pi }}{3}} \right)\).
 For \(\displaystyle \tan \theta =A,\,\,\theta ={{\tan }^{{1}}}A\,\,\,\left( {+\,\pi k} \right)\); this will find all the solutions. (For \(\cot \theta =A\), use \(\displaystyle {{\tan }^{{1}}}\left( {\frac{1}{A}} \right)\)). For example, in the interval \(\left[ {0,2\pi } \right)\), for \(\displaystyle \tan \theta =1,\,\,\theta ={{\tan }^{{1}}}\left( 1 \right)\,\approx .785\,\,\left( {\frac{\pi }{4}} \right)\), and also \(\displaystyle \theta ={{\tan }^{{1}}}\left( 1 \right)+\pi \,\approx 3.927\,\,\left( {\frac{{5\pi }}{4}} \right)\).
Remember that when the coefficient of the argument of the inverse trig functions isn’t 1, we need to divide the \(+\,2\pi k\) or \(+\,\pi k\) by this coefficient, since the period changes (see examples below).
If you have access to a graphing calculator, it’s usually easier to solve trig equations. We can put the lefthand part of the equation in \({{Y}_{1}}\) and the righthand part of the equation in \({{Y}_{2}}\) and solve for the intersection(s) between 0 and \(2\pi \). For solving over the real (general solutions), you can put the period of the trig function and then add the appropriate factors of \(\pi k\), \(2\pi k\), 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.
Typically, the default mode is radian mode, unless problem says “degrees”. When finding the intersection on the graphing calculator, use the TRACE and arrow buttons to move the cursor closer to the points of intersection, since the Intersect function will use the closest intersection.
For intervals of \(\left[ {0,2\pi } \right)\), use Xmin = 0, and Xmax = \(2\pi \). For general solutions (over the reals), use Xmin = 0 and Xmax = the period (such as \(\displaystyle \frac{2\pi }{5}\) 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 using both types of calendars:
Solving Trig Problems with a Calculator  
\(\displaystyle \sin x=\frac{3}{4}\text{ over Reals}\) (in degrees) Without Graphing: To get the second value of arcsin, subtract the first from 180 or \(\pi \). With Graphing: Use 0 for Xmin and 360 for Xmax: Add \(360{}^\circ k\) to answers: \(\displaystyle \left\{ \begin{array}{l}x\,x=48.6+360k\\\,\,\,\,\,\,\,\,\,\,=131.4+360k\end{array} \right\}\) 
\(\displaystyle \tan \left( {2\theta } \right)=\frac{1}{{\sqrt{6}}}\text{ on }\left[ {0,2\pi } \right)\) (in radians)
Without Graphing: To get the second value of arctan, add 180 or \(\pi \). For this problem, add \(\displaystyle \frac{\pi }{2}\) (the period) to get all the values between 0 and \(2\pi \). With Graphing: Use 0 for Xmin and \(2\pi \) for Xmax:
Get all intersections: \(\left\{ \begin{array}{l}\theta \theta =1.377,\,2.948\,\\\,\,\,\,\,\,\,\,=4.519,\,\,6.089\end{array} \right\}\) 
\(\displaystyle \cos \left( {3\theta 1} \right).8=0\text{ over Reals}\) Without Graphing: Work backwards to get the value of \(x\). To get the second value of arccos, we can take the negative cosine (same as subtracting the first from 360 or \(2\pi \)). For the answers .548 and .119, add \(\displaystyle \frac{{2\pi }}{3}k\) (period) to get all the solutions. With Graphing: (Easier!) Note that we use 0 for Xmin and \(\displaystyle \frac{{2\pi }}{3}\) for Xmax since \(\displaystyle \frac{{2\pi }}{3}\) is the period: Get both intersections and add factors of \(\displaystyle \frac{{2\pi }}{3}\) (the period): \(\displaystyle \left\{ \begin{align}\theta \,\theta &=.119+\frac{{2\pi k}}{3}\\\,\,\,\,\,\,\,\,\,\,\,\,&=\,.548+\frac{{2\pi k}}{3}\end{align} \right\}\) 
\(\displaystyle \cot \left( {\frac{2}{5}x\frac{\pi }{4}} \right)\text{ =10 over Reals}\)
Without Graphing: Work backwards to get the value of \(x\), using arctan of \(\displaystyle \frac{1}{{10}}\). For the answer 2.213, add \(\displaystyle \frac{{5\pi }}{2}k\) (period) to get all the solutions. With Graphing: Note that we use 0 for Xmin and \(\displaystyle \frac{{\,\pi }}{{\left( {\frac{2}{5}} \right)}}=\frac{{5\pi }}{2}\) for Xmax since \(\displaystyle \frac{{5\pi }}{2}\) is the period:
Add factors of \(\displaystyle \frac{{5\pi }}{2}\) (the period): \(\displaystyle \left\{ {x\,x=2.213+\frac{{5\pi k}}{2}} \right\}\) 
Solving Trig Systems of Equations
We learned how to solve systems of more complicated equations here in the Systems of nonLinear 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 Trig Systems of Equations  
\(\displaystyle \begin{array}{c}y\cos \left( x \right)=0\\y=\sin \left( x \right)\end{array}\)
\(\displaystyle \begin{array}{c}\sin x\cos x=0\\\sin x=\cos x\\\frac{{\sin x}}{{\cos x}}=1\\\,\tan x=1\end{array}\)
\(\displaystyle \begin{align}x=\frac{\pi }{4}+2\pi k\,\,\,\,&\,\,\,x=\frac{{3\pi }}{4}+2\pi k\\y=\sin \left( {\frac{\pi }{4}} \right)=\frac{{\sqrt{2}}}{2}\,\,\,\,\,&\,\,\,\,y=\sin \left( {\frac{{3\pi }}{4}} \right)=\frac{{\sqrt{2}}}{2}\end{align}\) \(\displaystyle \,\left( {\frac{\pi }{4}+2\pi k,\,\,\frac{{\sqrt{2}}}{2}} \right)\,\,\,\,\,\,\,\,\,\left( {\frac{{3\pi }}{4}+2\pi k,\,\,\frac{{\sqrt{2}}}{2}} \right)\) 
We’ll use substitution to get one equation, only with \(x\)’s. Since both sin and cos go on and on in the \(x\) direction, we have to include “\(\displaystyle +2\pi k\)” to our solutions.
(We have to be careful to separate the solutions we get to \(\displaystyle x=\frac{\pi }{4}+2\pi k\) and \(\displaystyle x=\frac{{3\pi }}{4}+2\pi k\), even though with tan, we can typically use \(\displaystyle x=\frac{\pi }{4}+\pi k\) for this solution. The reason we have to separate our answers like this is because we’ll have different answers for \(y\) for the two different places on the unit circle.)
Once we get the solutions for \(x\), we plug these back in either equation to get our \(y\) values. We see that the solutions are \(\displaystyle \left( {\frac{\pi }{4}+2\pi k,\frac{{\sqrt{2}}}{2}} \right)\) and \(\displaystyle \left( {\frac{{3\pi }}{4}+2\pi k,\frac{{\sqrt{2}}}{2}} \right)\). 
\(\displaystyle \begin{array}{c}\cos \left( x \right)+y=3\\y={{x}^{2}}\end{array}\)
Put in graphing calculator:
\(\displaystyle \begin{array}{l}{{Y}_{1}}=3\cos \left( x \right)\\{{Y}_{2}}={{x}^{2}}\end{array}\)

Let’s do this one on our graphing calculator (make sure your calculator is in radians). Solve for \(y\) in both cases, graph, and find the intersection. I made my window between \(2\pi \) and \(2\pi \) for \(x\).
Use the intersect feature on the calculator (2^{nd} trace, 5, enter, enter, enter) to find the intersection. (Use trace and arrow keys to get close to each intersection before using intersect). If we zoom out, we can see there are only two solutions. We see that the only solutions are \(\left( {1.795,3.222} \right)\) and \(\left( {1.795,3.222} \right)\).

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”:
Solving Trig Inequalities Graphically  Solving Trig Inequalities Algebraically 
\(6\left {\sin \theta } \right\le 3\,\,\,\,\,\,\text{over reals}\) \(\displaystyle \left {\sin \theta } \right\ge \frac{1}{2}\,\,\,\,\,\,\text{(switch signs!)}\) Graph both sides of the inequality to see where \(\left {\sin \left( x \right)} \right\) is greater than or equal to \(\displaystyle \frac{1}{2}\). Just graph between 0 and \(2\pi \) to see the points of intersection. We want above (including) \(\displaystyle \frac{1}{2}\), because of the \(\ge \): We see the solution is: \(\displaystyle \left( {\frac{\pi }{6}+2\pi k,\frac{{5\pi }}{6}+2\pi k} \right)\cup \left( {\frac{{7\pi }}{6}+2\pi k,\frac{{11\pi }}{6}+2\pi k} \right)\) 
Set everything to 0 and factor, so we can get our critical values:
\(\displaystyle 3{{\cot }^{2}}x+3\cot x\sqrt{3}\cot x<\sqrt{3};\,\,\,\,\,\,0\le x\le 2\pi \) \(\displaystyle \begin{array}{c}3{{\cot }^{2}}x+3\cot x\sqrt{3}\cot x\sqrt{3}<0\\3\cot x\left( {\cot x+1} \right)\sqrt{3}\left( {\cot x+1} \right)<0\\\left( {\cot x+1} \right)\left( {3\cot x\sqrt{3}} \right)<0\end{array}\) \(\displaystyle \,\cot x=1\,\,\,\,(x=\frac{{3\pi }}{4},\,\,\frac{{7\pi }}{4})\,\,\,\,\,\,\,\,\,\,\,\,\cot x=\frac{{\sqrt{3}}}{3}\,\,\,\,(x=\frac{\pi }{3},\,\,\frac{{4\pi }}{3})\)
Draw sign chart with critical values \(\displaystyle \frac{{3\pi }}{4},\frac{{7\pi }}{4},\frac{\pi }{3},\,\frac{{4\pi }}{3},\) and endpoints 0 and \(2\pi \), since we’re just going between 0 and \(2\pi \). Use open circles for the critical values since we have a \(<\). Then check each interval with a sample value in the last inequality above and see if we get a positive or negative value. We need to take values where the function is defined (avoid asymptotes), so let’s try \(\displaystyle \frac{\pi }{4}\) for the interval less than \(\displaystyle \frac{\pi }{3}\) for example: \(\displaystyle \left( {\cot \frac{\pi }{4}+1} \right)\left( {3\cot \frac{\pi }{4}\sqrt{3}} \right)\approx 2.5\), which is positive: We want the negative intervals, not including the critical values. We see the solution is: \(\displaystyle \left( {\frac{\pi }{3},\frac{{3\pi }}{4}} \right)\cup \left( {\frac{{4\pi }}{3},\frac{{7\pi }}{4}} \right)\). 
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