Solving Trigonometric Equations

Solving trigonometric equations is 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 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. In these cases, we want all solutions in the given interval.

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 that we always isolate the trig function, and some solutions may have none, or more than one solution. If there are multiply angles on the unit circle for that trig function, and an expression is involved, we may have to divide up the equation into two separate equations and solve each, like the example with $ \displaystyle \theta +\frac{\pi }{{18}}$.

If a square is involved, when we take the square root, we have to include both the positive and negative values. (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}}}$).

Note that sometimes you may have to solve using degrees $ \left[ {0,{{{360}}^{o}}} \right)$ instead of radians. 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\}$

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$ \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{align}2\sin \theta &=-1\\\sin \theta &=-\frac{1}{2}\end{align}$

$ \displaystyle \theta =\left\{ {\frac{{7\pi }}{6},\frac{{11\pi }}{6}} \right\}$

$ \displaystyle 4\sec \theta +7=-1$

 

$ \displaystyle \begin{align}4\sec \theta &=-8\\\sec \theta &=-2\,\,\,\left( {\cos \theta =-\frac{1}{2}} \right)\\\theta &=\left\{ {\frac{{2\pi }}{3},\frac{{4\pi }}{3}} \right\}\end{align}$

$ \displaystyle 2{{\cos }^{2}}\theta =1$

 

$ \displaystyle \begin{align}{{\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}\\\theta &=\left\{ {\frac{\pi }{4},\frac{{3\pi }}{4},\frac{{5\pi }}{4},\frac{{7\pi }}{4}} \right\}\end{align}$

$ \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 \frac{1}{4}+\sec \left( {x+\frac{\pi }{2}} \right)=\frac{3}{4}$

 

$ \displaystyle \begin{align}\text{sec}\left( {x+\frac{\pi }{2}} \right)=\frac{1}{2}\\\cos \left( {x+\frac{\pi }{2}} \right)=2\\x+\frac{\pi }{2}=\text{undefined}\\\text{No Solution, or }\emptyset \end{align}$

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$-values will be less (or to the left of) of those two 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, negative, or 0); sometimes this can be simplified.

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.

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.

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\}$

(also written as $ \displaystyle \left\{ {\theta |\theta =\frac{\pi }{6},\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\{ {x|x=\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\{ {x|x=\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\{ {x|x=\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{align}5\sec \theta &=-10\\\sec \theta &=-2\text{ }\left( {\cos \theta =-\frac{1}{2}} \right)\end{align}$

$ \displaystyle \left\{ \begin{array}{l}\theta |\theta =120{}^\circ +360{}^\circ k,\,\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,240{}^\circ +360{}^\circ k\end{array} \right\}$

$ \displaystyle 2\sin x=4\sin x$

 

$ \displaystyle \begin{array}{c}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{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\}$, or even $ \displaystyle \{\theta |\theta =\frac{\pi }{4}+\frac{\pi }{2}k\}$

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

 

$ \displaystyle \begin{align}x+\frac{\pi }{2}&=\frac{\pi }{3}+\pi k\,\,\,\,\text{(add}\,\pi k\,\text{for tan)}\\\,\,\,x&=-\frac{\pi }{6}+\pi k\end{align}$

Note that it’s better math “grammar” to write this without the negative angle:

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

$ \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\}\end{align}$

Watch domain restriction:

$ \displaystyle \sec x=\tan x\cdot {{\csc }^{2}}x$

 

$ \displaystyle \require {cancel} \frac{1}{{\cos \theta }}=\frac{{\cancel{{\sin \theta }}}}{{\cos \theta }}\cdot \frac{1}{{\cancel{{{{{\sin }}^{2}}{{\theta }_{{\sin \theta }}}}}}}$

$ \displaystyle \frac{1}{{\cos \theta }}=\frac{1}{{\sin \theta \cos \theta }}$

$ \displaystyle \begin{array}{c}\cos \theta =\sin \theta \cos \theta \,\,\,\,(\text{cross multiply)}\\\cos \theta -\sin \theta \cos \theta =0\\\cos \theta \left( {1-\sin \theta } \right)=0\\\cos \theta =0\,\,\,\,\,\,\,\,\sin \theta =1\end{array}$

$ \require {cancel} \displaystyle \left\{ {\theta |\cancel{{\theta =\frac{\pi }{2}+\pi k}}} \right\}$

Since $ \displaystyle \sec \left( {\frac{\pi }{2}} \right)$ is undefined, the solution is No Solution, or Ø.

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 $ \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$. Thus, 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 two 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)$, we’ll only get one solution instead of the normal two. And always check for extraneous solutions. Note that when we multiply or divide to get the variable by itself, we have to do the same with the “$ +2\pi k$” or “$ +\pi k$”.

Here are some problems: solve the following trig equations for 1) General Solutions, and 2) Solutions between $ \left[ {0,2\pi } \right)$ or $ \left[ {0,360{}^\circ } \right)$:

Solving Trig Problems with Multiple Angles 

$ \displaystyle \tan \left( {3\theta } \right)=\sqrt{3}$

 

General Solution:

$ \displaystyle \begin{align}3\theta =\frac{\pi }{3}+\pi k\,\,\,\,\\\left\{ {\theta |\theta =\frac{\pi }{9}+\frac{\pi }{3}k} \right\}\end{align}$

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Solution $ \left[ {0,2\pi } \right)$:

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$

 

General Solution:

$ \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\{ {x|x=\frac{{5\pi }}{{18}}+\frac{2}{3}\pi k,\,\,x=\frac{{7\pi }}{{18}}+\frac{2}{3}\pi k\,\,} \right\}$

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Solution $ \left[ {0,2\pi } \right)$:

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$ (degrees)

 

General Solution:

$ \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\{ {x|x=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\{ {x|x=\pm 270{}^\circ +2160{}^\circ k} \right\}$

————————————

Solution $ \left[ {0,360{}^\circ } \right)$:

Looking at the Unit Circle, the only solution that will work is:

$ \displaystyle x=\left\{ {270{}^\circ } \right\}$

$ \displaystyle 5{{\cos }^{2}}\left( {\frac{\theta }{3}} \right)=5$

 

General Solution:

$ \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}$

$ \displaystyle \begin{align}\frac{\theta }{3}=\pi k\\\left\{ {\theta |\theta =3\pi k} \right\}\end{align}$

————————————

Solution $ \left[ {0,2\pi } \right)$:

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$

 

General Solution:

$ \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}\end{align}$

$ \displaystyle \,2x=\frac{\pi }{4}+\pi k\,\,\,\,\,\,\,\,\,\,\,\,2x=\frac{{3\pi }}{4}+\pi k$

$ \displaystyle \left\{ {x|x=\frac{\pi }{8}+\frac{\pi }{2}k,\,\,\,x=\frac{{3\pi }}{8}+\frac{\pi }{2}k} \right\}$

————————————

Solution $ \left[ {0,2\pi } \right)$:

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 \sin \left( {\frac{\theta }{3}+\frac{\pi }{3}} \right)=\cos \left( {\frac{\theta }{3}+\frac{\pi }{3}} \right)$

 

General Solution:

$ \displaystyle \begin{align}\tan &\left( {\frac{\theta }{3}+\frac{\pi }{3}} \right)=1\\\,\,\,\frac{\theta }{3}+\frac{\pi }{3}&=\frac{\pi }{4}+\pi k\\\,\,\,\frac{\theta }{3}&=-\frac{\pi }{{12}}+\pi k\end{align}$

$ \displaystyle \,\left\{ {\theta |\theta =-\frac{\pi }{4}\,+3\pi k} \right\}$

 

————————————

Solution $ \left[ {0,2\pi } \right)$:

Looking at the Unit Circle, there are no solutions in $ \left[ {0,2\pi } \right)$, so $ \emptyset$.

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.

Here are some general hints when solving advanced trig equations:

  • When solving, simplify with identities first, if you can.
  • You can square each side, but don’t divide both sides by factors with variables, since you might be missing out on solutions. If you need to cross-multiply, or multiply both sides by what’s in a denominator (even when one side equals 0), make sure you’re not missing solutions. (It might be a good idea to see how many solutions there are in a graphing calculator if you can). And always check for extraneous solutions: solutions must work in the original equations, and denominators can’t be 0.
  • If you get answers for any trig function that has asymptotes (like tan), check for extraneous solutions (solutions that would be asymptotes).

Here are some examples, both solving on the interval 0 to $ 2\pi $ (or $ 360{}^\circ $) and over the reals. In the last problem, the answer ($ \displaystyle \theta = \frac{{\pi k}}{2}$) has to be “thrown out”, because of our domain restriction for cot (it falls on an asymptote); this 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 x-2\sin x=0\\2\sin x\left( {2\cos x-1} \right)=0\end{array}$

 

$ \displaystyle \begin{align}2\sin x=0\,\,\,\,\,\,&\,\,\,\,2\cos x-1=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 Solution (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 =150{}^\circ +180{}^\circ k\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\\\left\{ \begin{array}{l}\theta |\theta =30{}^\circ +180{}^\circ k,\,\,\theta =150{}^\circ +180{}^\circ k,\,\,\theta =180{}^\circ k\end{array} \right\}\end{array}$

$ \displaystyle \begin{array}{c}2{{\sec }^{2}}x-3\sec x=2\\\text{General Solution}\end{array}$

 

$ \displaystyle \begin{array}{c}2{{\sec }^{2}}x-3\sec x-2=0\\\left( {2\sec x+1} \right)\left( {\sec x-2} \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\{ {x|x=\frac{\pi }{3}+2\pi k,\,x=\frac{{5\pi }}{3}+2\pi k\,} \right\}$

$ \displaystyle \begin{array}{c}{{\tan }^{4}}x-4{{\tan }^{2}}x+3=0\\\text{General Solution}\end{array}$

 

$ \displaystyle \begin{array}{c}\left( {{{{\tan }}^{2}}x-3} \right)\left( {{{{\tan }}^{2}}x-1} \right)=0\\{{\tan }^{2}}x-3=0\,\,\,\,\,\,\,\,\,{{\tan }^{2}}x-1=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}x|x=\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\\\,\,\,\theta =\frac{{\pi k}}{2}\,\,\,\,\,\,\,\,\,&\,\,\,\,\,\,\,\theta =\frac{\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}$; it is extraneous. 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 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 use a scientific calculator, or graph the functions and find 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}\,} \right)$, and also $ \displaystyle \theta =\pi -{{\sin }^{{-1}}}\left( {.5\,} \right)\approx 2.618\,\,\left( {\frac{{5\pi }}{6}\,} \right)$.
  • 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)$, but be careful with the angles $ \displaystyle \frac{\pi }{2}+\pi k$, since the calculator shows undefined, instead of 0; you can use $ \displaystyle \cot =\frac{{\cos }}{{\sin }}$ instead). 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). Typically, the default mode is radian mode, unless problem says “degrees”.

If you have access to a graphing calculator, it’s usually easier to solve trig equations. We can put the left-hand part of the equation in $ {{Y}_{1}}$, the right-hand part of the equation in $ {{Y}_{2}}$, and solve for the intersection(s) between 0 and $ 2\pi $, or whatever the period when finding general solutions. (Use the trace feature and arrow keys to get close to each intersection, and then use the intersect feature (2nd trace, 5, enter, enter, enter) to find the intersection.) For the reciprocal functions, take the reciprocal of what’s on the right-hand side, and use the regular trig functions.

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. Then, for your answer, add the appropriate factors of $ \pi k$, $ 2\pi k$, or whatever the period of the function is.

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 calculators:

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, take the negative cos (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\}$

Note: We do have to be careful when using $ \frac{1}{{\tan \left( x \right)}}$ for $ \cot \left( x \right)$ in the calculator. For angles $ \frac{\pi }{2},\,\frac{{3\pi }}{2}$, the results won’t be correct; it shows an error, instead of 0 (try it!). It would be better to use $ \frac{{\cos \left( x \right)}}{{\sin \left( x \right)}}$ in this case.

Solving Trigonometric Systems of Equations

Systems of equations are needed when solving for more than one variable in equations. We learned how to solve systems of equations here in the the Systems of Linear Equations and Word Problems section, and systems of more complicated equations here in the Systems of non-Linear Equations section. Again, use either Substitution or Elimination, depending on what’s easier. Once we get the initial solution(s), we’ll can plug in a variable 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{{5\pi }}{4}+2\pi k\\y=\sin \left( {\frac{\pi }{4}} \right)=\frac{{\sqrt{2}}}{2}\,\,\,\,\,&\,\,\,\,y=\sin \left( {\frac{{5\pi }}{4}} \right)=-\frac{{\sqrt{2}}}{2}\end{align}$

$ \displaystyle \,\left( {\frac{\pi }{4}+2\pi k,\,\,\frac{{\sqrt{2}}}{2}} \right)\,\,\,\,\,\,\,\,\,\left( {\frac{{5\pi }}{4}+2\pi k,\,\,-\frac{{\sqrt{2}}}{2}} \right)$

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{{5\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{{5\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 trace feature and arrow keys to get close to each intersection, and then use the intersect feature (2nd trace, 5, enter, enter, enter) to find the intersection. 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 Trigonometric Inequalities

Sometimes you might be asked to solve a Trig Inequality. (Links to other types of Inequalities are found here).

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 \theta } \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 $:

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)$

$ \displaystyle 3{{\cot }^{2}}x+3\cot x-\sqrt{3}\cot x<\sqrt{3};\,\,\,\,\,\,0\le x<2\pi $

 

Set everything to 0 and factor, so we can get our critical values:

$ \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 a  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 that’s the interval we want. 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.

Take values where the function is defined (avoid asymptotes), so 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)$.

Practice these problems, and practice, practice, practice!


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On to Trigonometric Identities – you’re ready!