Trigonometry/Solving Trigonometric Equations

Trigonometric equations involve finding an unknown which is an argument to a trigonometric function.

Basic trigonometric equations

sin x = n

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

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

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

The solutions are then:

$x=\alpha +2k\pi$

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

Where $k$ is an integer.

In the cases when $n$ equals 1, 0 or -1 these solutions have simpler forms which are summarizied 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:

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

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

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

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

$x=\frac{4\pi }{3}(1+3k)$

Where $k$ is an integer.

cos x = n

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

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 the 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	$\begin{array}{c}x=\alpha +k\pi \\ \alpha \in [-\frac{\pi }{2};\frac{\pi }{2}]\end{array}$
$n=-1$	$x=-\begin{array}{c}\frac{\pi }{4}\end{array}+k\pi$
$n=0$	$x=k\pi$
$n=1$	$x=\begin{array}{c}\frac{\pi }{4}\end{array}+k\pi$

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	$\begin{array}{c}x=\alpha +k\pi \\ \alpha \in [0;\pi ]\end{array}$
$n=-1$	$x=-\begin{array}{c}\frac{3\pi }{4}\end{array}+k\pi$
$n=0$	$x=\begin{array}{c}\frac{\pi }{2}\end{array}+k\pi$
$n=1$	$x=\begin{array}{c}\frac{\pi }{4}\end{array}+k\pi$

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:

$\csc x=n\iff \frac{1}{\sin x}=n\iff \sin x=\frac{1}{n}$

$\sec x=n\iff \frac{1}{\cos x}=n\iff \cos x=\frac{1}{n}$

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:

$a\sin x+b\cos x=\sqrt{a^2+b^2}\sin (x+\alpha )$

$\alpha =\{\begin{array}{cc}{\tan }^{-1}(b/a),& \text{if }a>0\\ \pi +{\tan }^{-1}(b/a),& \text{if }a<0\end{array}$

The equation becomes:

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

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

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

For example we will solve:

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

In this case we have:

$a=1,b=-\sqrt{3}$

$\sqrt{a^2+b^2}=\sqrt{1^2+{(-\sqrt{3})}^2}=2$

$\alpha ={\tan }^{-1}(-\sqrt{3})=-\frac{\pi }{3}$

Apply the identity:

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

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

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

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

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

Where $k$ is an integer.