Precalculus - Trigonometric Equations

Introduction

• An equation involving one or more trigonometric functions (of some unknown angle) is called a trigonometric equation.
• A trigonometric identity is satisfied for every value of the unknown angle, while a trigonometric equation is satisfied only for some particular values of the unknown angle.
• The value of an unknown angle that satisfies a given trigonometric equation is called its root.
• $2\sin x-\sqrt{3}=0$, ${\cos }^{2}x-2\cos x+1=0$, etc are some examples of trigonometric equations.

Solving Trigonometric Equations

• A value of the unknown angle which satisfies a given trigonometric equation is called its solution.
• Since all the trigonometric functions are periodic and continue forever, trigonometric equations often have an infinite number of solutions unless the domain (angle values) is specified.
• There are basically two types of solutions:

(1) Principal solution or Primary Solution

(2) General Solution

(1) Principal solution or Primary Solution:

• We know that the values of $\sin x$ and $\cos x$ repeat after an interval of $2\mathrm{\pi }$ and the values of $\tan x$ repeat after an interval of $\mathrm{\pi }$.
• The solutions of trigonometric equations for which the variable (unknown angle) lies between 0 and $2\mathrm{\pi }$, are called Principal Solutions.

(2) General Solution:

• The complete set of values of the unknown angle satisfying a given trigonometric equation is called its general solution.
• The table shown below displays principal solutions and general solutions for some of the trigonometric equations.

Trigonometric Equation	Principal Solution	General Solution
$\sin x=0$	$x=0, \mathrm{\pi }, 2\mathrm{\pi }$	$x=k\mathrm{\pi }$, where $k\in Z$
$\sin x=1$	$x=\frac{\mathrm{\pi }}{2}$	$x=\frac{\mathrm{\pi }}{2}+2k\mathrm{\pi }$, where $k\in Z$
$\cos x=0$	$x=\frac{\mathrm{\pi }}{2}, \frac{3\mathrm{\pi }}{2}$	$x=\frac{\mathrm{\pi }}{2}+k\mathrm{\pi }$
$\cos x=1$	$x=0, 2\mathrm{\pi }$	$x=2k\mathrm{\pi }$

• If $sinx=\sin \alpha$, then $x=k\mathrm{\pi }+{\left(-1\right)}^{\mathrm{k}} \mathrm{\alpha }$, where $k\in Z$.
• If $\cos x=cos\alpha$, then $x=2k\mathrm{\pi }\pm \mathrm{\alpha }$.
• If $tanx=tan\alpha$, then $x=k\mathrm{\pi }\pm \mathrm{\alpha }$.

Solved Examples

Example 1: Find the principal solution of the given trigonometric equation.

$\mathbf{2}{\mathbf{cos}}^{\mathbf{2}}\mathit{x}\mathbf{+}\mathbf{3}\mathbf{cos}\mathit{x}\mathbf{-}\mathbf{2}\mathbf{=}\mathbf{0}$

Solution: $2{\cos }^{2}x+3\cos x-2=0$

Using quadratic formula:

$\cos x=\frac{-3\pm \sqrt{{\left(3\right)}^2-4\times 2\times \left(-2\right)}}{2\times \left(2\right)}$$=\frac{-3\pm \sqrt{9+16}}{4}$$=\frac{-3\pm \sqrt{25}}{4}$$=\frac{-3\pm 5}{4}$

$\Rightarrow \cos x=\frac{-3+5}{4}$   or   $\cos x=\frac{-3-5}{4}$

$\Rightarrow \cos x=\frac{2}{4}=\frac{1}{2}$   or   $\cos x=\frac{-8}{4}=-2$

$\Rightarrow \cos x=\frac{1}{2}$ or $-2$

Since the cosine function must range between − 1 and 1. The first answer, $\frac{1}{2}$ is a valid value. Thus, if k is an integer.

We know that $\cos \left(\frac{\mathrm{\pi }}{3}\right)=\frac{1}{2}$

Also, $cos\left(\frac{5\mathrm{\pi }}{3}\right)=$$\cos \left(2\mathrm{\pi }-\frac{\mathrm{\pi }}{3}\right)$$=\frac{1}{2}$

Therefore, the principal solutions are $\frac{\mathrm{\pi }}{3}$ and $\frac{5\mathrm{\pi }}{3}$.

 

Example 2: Find the general solution of the given trigonometric equation.

$\mathit{c}\mathit{o}\mathit{s}\mathbf{2}\mathit{x}\mathbf{+}\mathbf{3}\mathit{s}\mathit{i}\mathit{n}\mathit{x}\mathbf{+}\mathbf{1}\mathbf{=}\mathbf{0}$

Solution: $\cos 2x+3\sin x+1=0$

$\left(1-2{\sin }^{2}x\right)+3\sin x+1=0$

$-2{\sin }^{2}x+3\sin x+2=0$

$2{\sin }^{2}x-3\sin x-2=0$

Using quadratic formula:

$sinx=\frac{-3\pm \sqrt{{\left(3\right)}^2-4\times 2\times \left(-2\right)}}{2\times \left(2\right)}$$=\frac{-3\pm \sqrt{9+16}}{4}$$=\frac{-3\pm \sqrt{25}}{4}$$=\frac{-3\pm 5}{4}$

$\Rightarrow \sin x=\frac{-3+5}{4}$   or   $\sin x=\frac{-3-5}{4}$

$\Rightarrow \sin x=\frac{2}{4}=\frac{1}{2}$   or   $\sin x=\frac{-8}{4}=-2$

Since the sine function must range between − 1 and 1. The first answer, $\frac{1}{2}$ is a valid value. Thus, if k is an integer.

We know that $\sin \left(\frac{\mathrm{\pi }}{6}\right)=\frac{1}{2}$

Therefore, the general solution $x=k\mathrm{\pi }+{\left(-1\right)}^{\mathrm{k}} \left(\frac{\mathrm{\pi }}{6}\right)$, where $k\in Z$.

Cheat Sheet

• The solutions of trigonometric equations, for which its domain (angle-values) lies between 0 and 2π, are called principal solutions.
• The complete set of all the values of the unknown angle satisfying a trigonometric equation is called its general solution.
• If $sinx=\sin \alpha$, then $x=k\mathrm{\pi }+{\left(-1\right)}^{\mathrm{k}} \mathrm{\alpha }$, where $k\in Z$.
• If $\cos x=cos\alpha$, then $x=2k\mathrm{\pi }\pm \mathrm{\alpha }$.
• If $tanx=tan\alpha$, then $x=k\mathrm{\pi }\pm \mathrm{\alpha }$.

Blunder Areas

• It should be noted that not all trigonometric equations have solutions.
• In general trigonometric equations are usually solved using appropriate identities and algebraic manipulation.