Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 1 - Precalculus Review - 1.4 Trigonometric Functions - Exercises - Page 31: 46

Answer

$$ \frac{\pi}{6}, \frac{5 \pi}{6}, \frac{3 \pi}{2} $$

Work Step by Step

Since \begin{align*} \sin \theta&=\cos 2 \theta\\ &=\cos ^{2} \theta-\sin ^{2} \theta \\ &=1-\sin ^{2} \theta-\sin ^{2} \theta\\ &=1-2 \sin ^{2} \theta \end{align*} \begin{align*} 2\sin^2 \theta+\sin \theta-1&=0\\ (\sin \theta+1)(2\sin \theta-1)&=0 \end{align*} \begin{aligned} &\text { Now for } \sin \theta+1=0, \text { i.e. } \sin \theta=-1, \text { we get } \theta=\frac{3 \pi}{2}\\ &\text { And for } 2 \sin \theta-1=0, \text { we change it to } 2 \sin \theta=1\\ &\text { So, } \sin \theta=\frac{1}{2} \text { and we get } \theta=\frac{\pi}{6} \text { and } \frac{5 \pi}{6} \end{aligned} Then $$\theta \in\left\{\frac{\pi}{6}, \frac{5 \pi}{6}, \frac{3 \pi}{2}\right\}$$
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