## Precalculus (6th Edition) Blitzer

Published by Pearson

# Chapter 5 - Section 5.5 - Trigonometric Equations - Exercise Set - Page 704: 125

#### Answer

$0,\frac{2\pi}{3},\pi,\frac{4\pi}{3}$

#### Work Step by Step

Step 1. Factor the equation as $cos^2(x)(2cos(x)+1)-(2cos(x)+1)=(2cos(x)+1)(cos^2(x)-1)=0$ and use $cos^2(x)-1=-sin^2x$; we have $(2cos(x)+1)(sin^2(x))=0$ which gives $sin(x)=0$ and $cos(x)=-\frac{1}{2}$ Step 2. For $sin(x)=0$, the solutions in $[0,2\pi)$ are $x=0,\pi$ Step 3. For $cos(x)=-\frac{1}{2}$, we can find the solutions in $[0,2\pi)$ as $x=\frac{2\pi}{3},\frac{4\pi}{3}$ Step 4. The solutions to the original equation are $0,\frac{2\pi}{3},\pi,\frac{4\pi}{3}$

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