sin(x) = n

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

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

${\displaystyle \alpha =\arcsin(n)}$

The solutions are then:

${\displaystyle x=\alpha +2k\pi }$

${\displaystyle x=\pi -\alpha +2k\pi }$

Where ${\displaystyle k}$ is an integer.

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

For example, to solve:

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

First find ${\displaystyle \alpha }$ :

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

Then substitute in the formulae above:

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

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

Solving these linear equations for ${\displaystyle x}$ gives the final answer:

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

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

Where ${\displaystyle k}$ is an integer.

cos(x) = n

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

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

${\displaystyle \alpha =\arccos(n)}$

Then the solutions for ${\displaystyle x}$ are:

${\displaystyle x=\pm \alpha +2k\pi }$

Where ${\displaystyle k}$ is an integer.

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

tan(x) = n

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

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

${\displaystyle \alpha =\arctan(n)}$

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

${\displaystyle x=\alpha +k\pi }$

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

cot(x) = n

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

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

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

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

${\displaystyle x=\alpha +k\pi }$

When ${\displaystyle n}$ 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 ${\displaystyle \csc(x)=n}$ and ${\displaystyle \sec(x)=n}$ can be solved by transforming them to other basic equations:

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

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

a sin(x)+b cos(x) = c

To solve this equation we will use the identity:

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

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

The equation becomes:

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

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

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

For example we will solve:

${\displaystyle \sin(3x)-{\sqrt {3}}\cos(3x)=-{\sqrt {3}}}$

In this case we have:

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

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

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

Apply the identity:

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

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

So using the formulae for ${\displaystyle \sin(x)=n}$ the solutions to the equation are:

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

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

Where ${\displaystyle k}$ is an integer.