# 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

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

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

$\tan(x)=n$ :
| 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

$\cot(x)=n$ :
| 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.
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