Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 5 - Test - Page 709: 15


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

Work Step by Step

Step 1. Rewrite the function as $2(1-cos^2(x))+cos(x)=1$, we have $2cos^2(x)-cos(x)-1=0$ Step 2. Factor the left side to get $(2cos(x)+1)(cos(x)-1)=0$ Step 3. For $cos(x)=1$, we have $x=0$ Step 4. For $cos(x)=-\frac{1}{2}$, we have $x=\frac{2\pi}{3},\frac{4\pi}{3} $ Step 5. We have the solution set as $\{0, \frac{2\pi}{3},\frac{4\pi}{3} \}$
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