## Trigonometry (11th Edition) Clone

Published by Pearson

# Chapter 6 - Inverse Circular Functions and Trigonometric Equations - Section 6.2 Trigonometric Equations I - 6.2 Exercises - Page 273: 21

#### Answer

The solution set to this problem is $$\{\frac{\pi}{4},\frac{2\pi}{3},\frac{5\pi}{4},\frac{5\pi}{3}\}$$

#### Work Step by Step

$$(\cot x-1)(\sqrt3\cot x+1)=0$$ over interval $[0,2\pi)$ 1) Solve the equation: $$(\cot x-1)(\sqrt3\cot x+1)=0$$ $$\cot x-1=0\hspace{1cm}\text{or}\hspace{1cm}\sqrt3\cot x+1=0$$ $$\cot x=1\hspace{1cm}\text{or}\hspace{1cm}\cot x=-\frac{1}{\sqrt3}=-\frac{\sqrt3}{3}$$ 2) Apply the inverse function: - $\cot x=1$: Over the interval $[0,2\pi)$, there are 2 values whose cotangent equals $1$, which are $\frac{\pi}{4}$ and $\frac{5\pi}{4}$ - $\cot x=-\frac{\sqrt3}{3}$: Over the interval $[0,2\pi)$, there are 2 values whose cotangent equals $-\frac{\sqrt3}{3}$, which are $\frac{2\pi}{3}$ and $\frac{5\pi}{3}$ Therefore, $$x\in\{\frac{\pi}{4},\frac{2\pi}{3},\frac{5\pi}{4},\frac{5\pi}{3}\}$$ In other words, the solution set to this problem is $$\{\frac{\pi}{4},\frac{2\pi}{3},\frac{5\pi}{4},\frac{5\pi}{3}\}$$

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.