Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

Published by Pearson
ISBN 10: 0-32193-104-1
ISBN 13: 978-0-32193-104-7

Chapter 6 - Analytic Trigonometry - Chapter Review - Review Exercises - Page 528: 74

Answer

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

Work Step by Step

1. $sin(2\theta)=\sqrt 2cos(\theta) \Longrightarrow 2sin(\theta)cos(\theta)=\sqrt 2cos(\theta) \Longrightarrow cos(\theta)=0\ or\ sin(\theta)=\frac{\sqrt 2}{2}$, 2. For $cos(\theta)=0$, we have $\theta=k\pi+\frac{\pi}{2}$ 3. For $sin(\theta)=\frac{\sqrt 2}{2}$, we have $\theta=2k\pi+\frac{\pi}{4}$ or $\theta=2k\pi+\frac{3\pi}{4}$ 4. Within $[0,2\pi)$, we have $\theta=\frac{\pi}{4},\frac{\pi}{2},\frac{3\pi}{4},\frac{3\pi}{2}$
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