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

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

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:

$\alpha =\arcsin(n)$

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 summarized in the table on the right.

For example, to solve:

$\sin {\bigl (}{\tfrac {x}{2}}{\bigr )}={\frac {\sqrt {3}}{2}}$

First find $\alpha$ :

$\alpha =\arcsin {\bigl (}{\tfrac {\sqrt {3}}{2}}{\bigr )}={\frac {\pi }{3}}$

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

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

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:

$\alpha =\arccos(n)$

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:

$\alpha =\arctan(n)$

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:

$\alpha =\operatorname {arccot}(n)$

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:

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

$\alpha ={\begin{cases}\arctan {\bigl (}{\frac {b}{a}}{\bigr )},&{\mbox{if }}a>0\\\pi +\arctan {\bigl (}{\frac {b}{a}}{\bigr )},&{\mbox{if }}a<0\end{cases}}$

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:

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

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

$\alpha =\arctan {\Big (}-{\sqrt {3}}{\Big )}=-{\frac {\pi }{3}}$

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:

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

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

Where $k$ is an integer.