Precalculus (6th Edition) Blitzer

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

Chapter 5 - Review Exercises - Page 709: 61


$0,\pi,\frac{7\pi}{6}, \frac{11\pi}{6}$.

Work Step by Step

Step 1. Rewrite the equation as $(1-2sin^2x)-sin(x)=1$ or $2sin^2x+sin(x)=sin(x)(2sin(x)+1)=0$ Step 2. For $sin(x)=0$, we have solutions in $[0,2\pi)$ as $x=0,\pi$ Step 3. For $sin(x)=-\frac{1}{2}$, we have solutions in $[0,2\pi)$ as $x=\frac{7\pi}{6}, \frac{11\pi}{6}$ Step 4. Thus, we have all the solutions in $[0,2\pi)$ as $x=0,\pi,\frac{7\pi}{6}, \frac{11\pi}{6}$.
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