# How to Solve Basic Trigonometric Equations

When solving trigonometric equations, it’s very important to include all the solutions! That means you have to be aware of the period of the function you are working with and how it influences the number of solutions.

Rule

### BasicTrigonometricEquations

When $a,b\in \phantom{\rule{-0.17em}{0ex}}\left[-1,1\right]$ and $n\in ℤ$, the following is true:

$\begin{array}{llll}\hfill \mathrm{sin}x=a⇒x& =u+n\cdot 2\pi \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill x& =\pi -u+n\cdot 2\pi \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill \mathrm{cos}x=b⇒x& =u+n\cdot 2\pi \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill x& =-u+n\cdot 2\pi \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill \mathrm{tan}x=c⇒x& =u+n\cdot \pi \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

$\begin{array}{llll}\hfill \mathrm{sin}x=a⇒x& =u+n\cdot 2\pi \phantom{\rule{1em}{0ex}}\vee \phantom{\rule{1em}{0ex}}x=\pi -u+n\cdot 2\pi \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill \mathrm{cos}x=b⇒x& =u+n\cdot 2\pi \phantom{\rule{1em}{0ex}}\vee \phantom{\rule{1em}{0ex}}x=-u+n\cdot 2\pi \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill \mathrm{tan}x=c⇒x& =u+n\cdot \pi \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

Here, $u$ represents the number you find for ${\mathrm{sin}}^{-1}\left(a\right)$, ${\mathrm{cos}}^{-1}\left(b\right)$ or ${\mathrm{tan}}^{-1}\left(c\right)$.

Note! It’s very important to check which values $x$ is allowed to have. It varies from problem to problem, and influences which values for $n$ you can use in your answer.

Example 1

Solve the equation $4\mathrm{cos}\phantom{\rule{-0.17em}{0ex}}\left(\pi x+\frac{2\pi }{3}\right)=2$ for $x\in \phantom{\rule{-0.17em}{0ex}}\left(0,2\pi \right)$

You begin by transforming the equation to get the $\mathrm{cos}$-term on its own: $\begin{array}{llll}\hfill 4\mathrm{cos}\phantom{\rule{-0.17em}{0ex}}\left(\pi x+\frac{2\pi }{3}\right)& =2,\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill \mathrm{cos}\phantom{\rule{-0.17em}{0ex}}\left(\pi x+\frac{2\pi }{3}\right)& =\frac{1}{2}.\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

This has the solutions $\begin{array}{lll}\hfill \pi {x}_{1}+\frac{2\pi }{3}& =\frac{\pi }{3}+n\cdot 2\pi ,\phantom{\rule{2em}{0ex}}& \hfill \text{(1)}\phantom{\rule{0.33em}{0ex}}\\ \hfill \pi {x}_{2}+\frac{2\pi }{3}& =-\frac{\pi }{3}+n\cdot 2\pi .\phantom{\rule{2em}{0ex}}& \hfill \text{(2)}\phantom{\rule{0.33em}{0ex}}\end{array}$

First, you continue with (1): $\begin{array}{llll}\hfill \pi {x}_{1}+\frac{2\pi }{3}& =\frac{\pi }{3}+n\cdot 2\pi \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill \pi {x}_{1}& =-\frac{\pi }{3}+n\cdot 2\pi \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill {x}_{1}& =-\frac{1}{3}+2n\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

Then you continue with (2): $\begin{array}{llll}\hfill \pi {x}_{2}+\frac{2\pi }{3}& =-\frac{\pi }{3}+n\cdot 2\pi \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill \pi {x}_{2}& =-\pi +n\cdot 2\pi \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill {x}_{2}& =-1+2n\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

The problem tells you to find all the solutions that are in the interval $x\in \phantom{\rule{-0.17em}{0ex}}\left(0,2\pi \right)$. You find these by considering ${x}_{1}$ and ${x}_{2}$ with respect to that interval.

Look at ${x}_{1}=-\frac{1}{3}+2n$ first. If you insert $n=1$, you get

 ${x}_{1}=-\frac{1}{3}+2\cdot 1=\frac{5}{3},$

which is in the interval. When you check $n=2$, you get

 ${x}_{1}=-\frac{1}{3}+2\cdot 2=\frac{11}{3}\in \phantom{\rule{-0.17em}{0ex}}\left(0,2\pi \right).$

Then you check $n=3$,

 ${x}_{1}=-\frac{1}{3}+2\cdot 3=\frac{17}{3},$

which is also in the interval. Now you notice that if you check $n=4$, the answer will be outside the interval $\phantom{\rule{-0.17em}{0ex}}\left(0,2\pi \approx 6.28\right)$. That means you have found all the solutions for ${x}_{1}$.

Now you have to do the same for ${x}_{2}=-1+2n$. The values still have to be in the interval for them to be a part of the solution. That gives you $\begin{array}{llll}\hfill n=1⇒{x}_{2}& =-1+2\cdot 1=1\in \phantom{\rule{-0.17em}{0ex}}\left(0,2\pi \right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill n=2⇒{x}_{2}& =-1+2\cdot 2=3\in \phantom{\rule{-0.17em}{0ex}}\left(0,2\pi \right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill n=3⇒{x}_{2}& =-1+2\cdot 3=5\in \phantom{\rule{-0.17em}{0ex}}\left(0,2\pi \right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill n=4⇒{x}_{2}& =-1+2\cdot 4=7\notin \phantom{\rule{-0.17em}{0ex}}\left(0,2\pi \right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

As you can see, the last value is outside the interval. That means the solutions in the interval $\phantom{\rule{-0.17em}{0ex}}\left(0,2\pi \right)$ are:

 $x\in \phantom{\rule{-0.17em}{0ex}}\left\{1,\frac{5}{3},3,\frac{11}{3},5,\frac{17}{3}\right\}.$

Example 2

Solve the equation $\mathrm{sin}\phantom{\rule{-0.17em}{0ex}}\left(2x-\frac{\pi }{3}\right)=\frac{1}{2}$ for $x\in \phantom{\rule{-0.17em}{0ex}}\left[0,2\pi \right)$

The basic equation

 $\mathrm{sin}\phantom{\rule{-0.17em}{0ex}}\left(2x-\frac{\pi }{3}\right)=\frac{1}{2}$

has the solutions $\begin{array}{lll}\hfill 2{x}_{1}-\frac{\pi }{3}& ={\mathrm{sin}}^{-1}\phantom{\rule{-0.17em}{0ex}}\left(\frac{1}{2}\right)+n\cdot 2\pi ,\phantom{\rule{2em}{0ex}}& \hfill \text{(3)}\phantom{\rule{0.33em}{0ex}}\\ \hfill 2{x}_{2}-\frac{\pi }{3}& =\pi -{\mathrm{sin}}^{-1}\phantom{\rule{-0.17em}{0ex}}\left(\frac{1}{2}\right)+n\cdot 2\pi .\phantom{\rule{2em}{0ex}}& \hfill \text{(4)}\phantom{\rule{0.33em}{0ex}}\end{array}$

First, you continue working on (3): $\begin{array}{llll}\hfill 2{x}_{1}-\frac{\pi }{3}& =\frac{\pi }{6}+n\cdot 2\pi \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill 2{x}_{1}& =\frac{\pi }{2}+n\cdot 2\pi \phantom{\rule{1em}{0ex}}|:2\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill {x}_{1}& =\frac{\pi }{4}+n\cdot \pi \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

Then you work on (4): $\begin{array}{llll}\hfill 2{x}_{2}-\frac{\pi }{3}& =\pi -\frac{\pi }{6}+n\cdot 2\pi \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill 2{x}_{2}& =\frac{7\pi }{6}+n\cdot 2\pi \phantom{\rule{1em}{0ex}}|:2\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill x& =\frac{7\pi }{12}+n\cdot \pi \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

Now you have to find the solutions from ${x}_{1}$ and ${x}_{2}$. The values have to be in the interval $x\in \phantom{\rule{-0.17em}{0ex}}\left[0,2\pi \right)$ for them to be one of the solutions. For ${x}_{1}=\frac{\pi }{4}+2\pi$, you have $\begin{array}{llll}\hfill n=0⇒{x}_{1}& =\frac{\pi }{4}+0=\frac{\pi }{4},\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill n=1⇒{x}_{1}& =\frac{\pi }{4}+1\pi =\frac{5\pi }{4},\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill n=2⇒{x}_{1}& =\frac{\pi }{4}+2\pi =\frac{9\pi }{4},\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

where $\frac{9\pi }{4}$ is outside the interval. When you check ${x}_{2}=\frac{7\pi }{12}+n\pi$, you get $\begin{array}{llll}\hfill n=0⇒{x}_{2}& =\frac{7\pi }{12}+0=\frac{7\pi }{12},\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill n=1⇒{x}_{2}& =\frac{7\pi }{12}+1\pi =\frac{19\pi }{12},\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill n=2⇒{x}_{2}& =\frac{7\pi }{12}+2\pi =\frac{31\pi }{12},\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

where $\frac{31\pi }{12}$ is outside the interval. That means solutions in the interval $\phantom{\rule{-0.17em}{0ex}}\left[0,2\pi \right)$ are:

 $x\in \phantom{\rule{-0.17em}{0ex}}\left\{\frac{\pi }{4},\frac{7\pi }{12},\frac{5\pi }{4},\frac{19\pi }{12}\right\}.$

Example 3

Solve the equation $3\mathrm{tan}\phantom{\rule{-0.17em}{0ex}}\left(3x+\frac{5\pi }{6}\right)=3$ for $x\in ℝ$

You solve the trigonometric equation for $x$:

$\begin{array}{llll}\hfill 23\mathrm{tan}\phantom{\rule{-0.17em}{0ex}}\left(3x+\frac{5\pi }{6}\right)& =3\phantom{\rule{2em}{0ex}}|:3\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill \mathrm{tan}\phantom{\rule{-0.17em}{0ex}}\left(3x+\frac{5\pi }{6}\right)& =1\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill 3x+\frac{5\pi }{6}& ={\mathrm{tan}}^{-1}\left(1\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill 3x+\frac{5\pi }{6}& =\frac{\pi }{4}+n\pi \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill 3x& =\frac{\pi }{4}-\frac{5\pi }{6}+n\pi \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill 3x& =\frac{3\pi -10\pi }{12}+n\pi \phantom{\rule{1em}{0ex}}|:3\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill x& =-\frac{7\pi }{36}+\frac{n\pi }{3}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

$\begin{array}{llllllll}\hfill 3\mathrm{tan}\phantom{\rule{-0.17em}{0ex}}\left(3x+\frac{5\pi }{6}\right)& =3\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{1em}{0ex}}|:3\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill \mathrm{tan}\phantom{\rule{-0.17em}{0ex}}\left(3x+\frac{5\pi }{6}\right)& =1\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill 3x+\frac{5\pi }{6}& ={\mathrm{tan}}^{-1}\left(1\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill 3x+\frac{5\pi }{6}& =\frac{\pi }{4}+n\pi \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill 3x& =\frac{\pi }{4}-\frac{5\pi }{6}+n\pi \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill 3x& =\frac{3\pi -10\pi }{12}+n\pi \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{1em}{0ex}}|:3\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill x& =-\frac{7\pi }{36}+\frac{n\pi }{3}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

The solution to the equation is $x=-\frac{7\pi }{36}+\frac{n\pi }{3}$ for every $n\in ℤ$. Because $x$ can be any real number in this example ($x\in ℝ$), there is no specific interval you need to check $n$-values for. That means the solution is general.