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.

Contents

Basic trigonometric equationsEdit

sin(x) = nEdit

Sin unit circle.svg
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{\sqrt3}{2}

First find \alpha :

\alpha=\arcsin\bigl(\tfrac{\sqrt3}{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) = nEdit

Cos unit circle.svg
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) = nEdit

Tan unit circle.svg
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) = nEdit

Cot unit circle.svg
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=\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) = nEdit

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 examplesEdit

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) = cEdit

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)-\sqrt3\cos(3x)=-\sqrt3

In this case we have:

a=1,b=-\sqrt3
\sqrt{a^2+b^2}=\sqrt{1^2+\Big(-\sqrt3\Big)^2}=2
\alpha=\arctan\Big(-\sqrt3\Big)=-\frac{\pi}{3}

Apply the identity:

2\sin\left(3x-\frac{\pi}{3}\right)=-\sqrt3
\sin\left(3x-\frac{\pi}{3}\right)=-\frac{\sqrt3}{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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