Algebra and Trigonometry 10th Edition

Published by Cengage Learning
ISBN 10: 9781337271172
ISBN 13: 978-1-33727-117-2

Chapter 7 - Review Exercises - Page 552: 37


$x=\frac{\pi}{8}$ $x=\frac{3\pi}{8}$ $x=\frac{9\pi}{8}$ $x=\frac{11\pi}{8}$

Work Step by Step

$2~sin~2x-\sqrt 2=0$ $2~sin~2x=\sqrt 2$ $sin~2x=\frac{\sqrt 2}{2}$ $2x=\frac{\pi}{4}$ or $2x=\frac{3\pi}{4}$ (with $0\lt2x\lt2\pi$) If $x$ is in the interval $[0,2\pi)$, then $2x$ is in the interval $[0,4\pi)$ The period of $sin~θ$ is $2\pi$. So, we can add $2\pi$ to each answer above: $2x=\frac{\pi}{4}+2\pi=\frac{9\pi}{4}$ $2x=\frac{3\pi}{4}+2\pi=\frac{11\pi}{4}$ Finally: $2x=\frac{\pi}{4}$ $x=\frac{\pi}{8}$ $2x=\frac{3\pi}{4}$ $x=\frac{3\pi}{8}$ $2x=\frac{9\pi}{4}$ $x=\frac{9\pi}{8}$ $2x=\frac{11\pi}{4}$ $x=\frac{11\pi}{8}$
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