Trigonometry/Solving Trigonometric Equations

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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[edit]

sin x = n[edit]

Sin unit circle.png
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 first find an \alpha 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 summarized 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} + 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.

cos x = n[edit]

Cos unit circle.png
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 one 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[edit]

Tan unit circle.png
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=\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[edit]

Cot unit circle.png
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=\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[edit]

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[edit]

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[edit]

To solve this equation we will use the identity:

a \sin x + b \cos x = \sqrt{a^2 + b^2} \sin \left(x + \alpha\right)
\alpha = \begin{cases} \tan^{-1} \left(b / a\right), & \mbox{if } a > 0 \\ \pi + \tan^{-1} \left(b / a\right), & \mbox{if } a < 0 \end{cases}

The equation becomes:

\sqrt{a^2 + b^2} \sin \left(x + \alpha\right) = c
\sin{\left(x + \alpha\right)} = \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 3 x - \sqrt{3}\cos 3 x = -\sqrt{3}

In this case we have:

a = 1, b = -\sqrt{3}
\sqrt{a^2+b^2} = \sqrt{1^2 + \left(-\sqrt{3}\right)^2} = 2
\alpha = \tan^{-1} \left(-\sqrt{3}\right) = -\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} + 2 k \pi \Leftrightarrow x = \frac{2 k \pi}{3}
3x - \frac{\pi}{3} = \pi + \frac{\pi}{3} + 2 k \pi \Leftrightarrow x = \frac{\pi}{9}\left(6k + 5\right)

Where k is an integer.