Trigonometry/Solving Trigonometric Equations

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

[edit] Basic trigonometric equations

[edit] 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 need to find an $α$ 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 summarizied in the table on the right.

For example, to solve:

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

First find $α$ :

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

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.

[edit] 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 the angle $α$ 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.

[edit] 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 $α$ such that:

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

After finding $α$ , 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.

[edit] 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 $α$ such that:

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

After finding $α$ , 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.

[edit] 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:

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

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

[edit] 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.