# 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\mathrm{sin}x-\sqrt{3}=0$${\mathrm{cos}}^{2}x-2\mathrm{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 $\mathrm{sin}x$ and $\mathrm{cos}x$ repeat after an interval of $2\mathrm{\pi }$ and the values of $\mathrm{tan}x$ 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 $\mathrm{sin}x=0$ $x=k\mathrm{\pi }$, where $k\in Z$ $\mathrm{sin}x=1$ $x=\frac{\mathrm{\pi }}{2}$ $x=\frac{\mathrm{\pi }}{2}+2k\mathrm{\pi }$, where $k\in Z$ $\mathrm{cos}x=0$ $x=\frac{\mathrm{\pi }}{2}+k\mathrm{\pi }$ $\mathrm{cos}x=1$ $x=2k\mathrm{\pi }$
• If $sinx=\mathrm{sin}\alpha$, then , where $k\in Z$.
• If $\mathrm{cos}x=cos\alpha$, then $x=2k\mathrm{\pi }±\mathrm{\alpha }$.
• If $tanx=tan\alpha$, then $x=k\mathrm{\pi }±\mathrm{\alpha }$.

## Solved Examples

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

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

Solution: $2{\mathrm{cos}}^{2}x+3\mathrm{cos}x-2=0$

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

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

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

$⇒\mathrm{cos}x=\frac{1}{2}$ 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 $\mathrm{cos}\left(\frac{\mathrm{\pi }}{3}\right)=\frac{1}{2}$

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

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

Solution: $\mathrm{cos}2x+3\mathrm{sin}x+1=0$

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

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

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

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

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

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

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

Therefore, the general solution , 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 $sinx=\mathrm{sin}\alpha$, then , where $k\in Z$.
• If $\mathrm{cos}x=cos\alpha$, then $x=2k\mathrm{\pi }±\mathrm{\alpha }$.
• If $tanx=tan\alpha$, then $x=k\mathrm{\pi }±\mathrm{\alpha }$.

## 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.