# Trigonometry/Solving Trigonometric Equations

Trigonometric equations are equations including trigonometric functions. If they have only such functions and constants, then the solution involves finding an unknown which is an argument to a trigonometric function.

## Basic trigonometric equations

### sin x = n

$n\,\!$ $\sin x=n\,\!$ $\left|n\right|<1$ $\begin{matrix}x=\alpha + 2 k \pi \\ x=\pi - \alpha + 2 k \pi \\ \alpha \in \left[-\frac{\pi}{2};\frac{\pi}{2}\right]\end{matrix}$ $n=-1\,\!$ $x=-\begin{matrix}\frac{\pi}{2}\end{matrix}+2k\pi$ $n=0\,\!$ $x=k\pi\,\!$ $n=1\,\!$ $x=\begin{matrix}\frac{\pi}{2}\end{matrix}+2k\pi$ $\left|n\right|>1$ $x\in\varnothing$

The equation $\sin x=n$ has solutions only when $n$ is within the interval [-1; 1]. If $n$ is within this interval, then we first find an $\alpha$ such that:

$\alpha=\sin^{-1} n\,\!$

The solutions are then:

$x=\alpha + 2 k \pi\,\!$
$x=\pi - \alpha + 2 k \pi\,\!$

Where $k$ is an integer.

In the cases when $n$ equals 1, 0 or -1 these solutions have simpler forms which are summarized in the table on the right.

For example, to solve:

$\sin \frac{x}{2}=\frac{\sqrt{3}}{2}$

First find $\alpha$:

$\alpha=\sin^{-1}{\frac{\sqrt{3}}{2}}=\frac{\pi}{3}$

Then substitute in the formulae above:

$\frac{x}{2}=\frac{\pi}{3} + 2 k \pi$
$\frac{x}{2}=\pi - \frac{\pi}{3} + 2 k \pi$

Solving these linear equations for $x$ gives the final answer:

$x=\frac{2\pi}{3}\left(1+6k\right)$
$x=\frac{4\pi}{3}\left(1+3k\right)$

Where $k$ is an integer.

### cos x = n

$n\,\!$ $\cos x=n\,\!$ $\left|n\right|<1$ $\begin{matrix}x=\pm\alpha + 2 k \pi \\ \alpha \in \left[0;\pi\right]\end{matrix}$ $n=-1\,\!$ $x=\pi+2k\pi\,\!$ $n=0\,\!$ $x=\begin{matrix}\frac{\pi}{2}\end{matrix}+k\pi\,\!$ $n=1\,\!$ $x=2k\pi\,\!$ $\left|n\right|>1$ $x\in\varnothing$

Like the sine equation, an equation of the form $\cos x=n$ only has solutions when n is in the interval [-1; 1]. To solve such an equation we first find one angle $\alpha$ such that:

$\alpha=\cos^{-1}n\,\!$

Then the solutions for $x$ are:

$x=\pm\alpha+2k\pi \, \!$

Where $k$ is an integer.

Simpler cases with $n$ equal to 1, 0 or -1 are summarized in the table on the right.

### tan x = n

$n\,\!$ $\tan x=n\,\!$ General case $\begin{matrix}x=\alpha + k\pi \\ \alpha \in \left[-\frac{\pi}{2};\frac{\pi}{2}\right]\end{matrix}$ $n=-1\,\!$ $x=-\begin{matrix}\frac{\pi}{4}\end{matrix}+k\pi\,\!$ $n=0\,\!$ $x=k\pi\,\!$ $n=1\,\!$ $x=\begin{matrix}\frac{\pi}{4}\end{matrix}+k\pi\,\!$

An equation of the form $\tan x=n$ has solutions for any real $n$. To find them we must first find an angle $\alpha$ such that:

$\alpha=\tan^{-1}n\,\!$

After finding $\alpha$, the solutions for $x$ are:

$x=\alpha+k\pi\,\!$

When $n$ equals 1, 0 or -1 the solutions have simpler forms which are shown in the table on the right.

### cot x = n

$n\,\!$ $\cot x=n\,\!$ General case $\begin{matrix}x=\alpha + k\pi \\ \alpha \in \left[0;\pi\right]\end{matrix}$ $n=-1\,\!$ $x=-\begin{matrix}\frac{3\pi}{4}\end{matrix}+k\pi\,\!$ $n=0\,\!$ $x=\begin{matrix}\frac{\pi}{2}\end{matrix}+k\pi\,\!$ $n=1\,\!$ $x=\begin{matrix}\frac{\pi}{4}\end{matrix}+k\pi\,\!$

The equation $\cot x=n$ has solutions for any real $n$. To find them we must first find an angle $\alpha$ such that:

$\alpha=\cot^{-1}n\,\!$

After finding $\alpha$, the solutions for $x$ are:

$x=\alpha+k\pi\,\!$

When $n$ equals 1, 0 or -1 the solutions have simpler forms which are shown in the table on the right.

### csc x = n and sec x = n

The trigonometric equations csc x = n and sec x = n can be solved by transforming them to other basic equations:

$\csc x=n\Leftrightarrow\frac{1}{\sin x}=n\Leftrightarrow\sin x=\frac{1}{n}$
$\sec x=n\Leftrightarrow\frac{1}{\cos x}=n\Leftrightarrow\cos x=\frac{1}{n}$

## Further examples

Generally, to solve trigonometric equations we must first transform them to a basic trigonometric equation using the trigonometric identities. This sections lists some common examples.

### a sin x + b cos x = c

To solve this equation we will use the identity:

$a \sin x + b \cos x = \sqrt{a^2 + b^2} \sin \left(x + \alpha\right)$
$\alpha = \begin{cases} \tan^{-1} \left(b / a\right), & \mbox{if } a > 0 \\ \pi + \tan^{-1} \left(b / a\right), & \mbox{if } a < 0 \end{cases}$

The equation becomes:

$\sqrt{a^2 + b^2} \sin \left(x + \alpha\right) = c$
$\sin{\left(x + \alpha\right)} = \frac{c}{\sqrt{a^2 + b^2}}$

This equation is of the form $\sin x = n$ and can be solved with the formulae given above.

For example we will solve:

$\sin 3 x - \sqrt{3}\cos 3 x = -\sqrt{3}$

In this case we have:

$a = 1, b = -\sqrt{3}$
$\sqrt{a^2+b^2} = \sqrt{1^2 + \left(-\sqrt{3}\right)^2} = 2$
$\alpha = \tan^{-1} \left(-\sqrt{3}\right) = -\frac{\pi}{3}$

Apply the identity:

$2 \sin \left(3x - \frac{\pi}{3}\right) = -\sqrt{3}$
$\sin \left(3x - \frac{\pi}{3}\right) = -\frac{\sqrt{3}}{2}$

So using the formulae for $\sin x = n$ the solutions to the equation are:

$3x - \frac{\pi}{3} = -\frac{\pi}{3} + 2 k \pi \Leftrightarrow x = \frac{2 k \pi}{3}$
$3x - \frac{\pi}{3} = \pi + \frac{\pi}{3} + 2 k \pi \Leftrightarrow x = \frac{\pi}{9}\left(6k + 5\right)$

Where $k$ is an integer.