Calculus (3rd Edition)

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

Chapter 1 - Precalculus Review - Chapter Review Exercises - Page 37: 47


$$ x= \frac{\pi}{2}, \frac{3 \pi}{2}, \frac{7 \pi}{6}, \frac{11 \pi}{6}$$

Work Step by Step

We are given $$\sin 2 x+\cos x=0 \text { for } 0 \leq x<2 \pi$$ We factor the left side (and use the double angle formula) as follows: \begin{align*} \sin 2 x+\cos x&=0\\ 2\sin x\cos x+\cos x&=0\\ \cos x(2\sin x+1)&=0 \end{align*} Now we set the factors equal to zero $$ \cos x=0\ \ \ \ \Rightarrow \ \ \ x= \frac{\pi}{2}, \frac{3 \pi}{2}$$ or $$ 2\sin x+1=0\ \ \ \ \Rightarrow \ \ \ x=\frac{7 \pi}{6}, \frac{11 \pi}{6}$$ Hence the solutions are: $$ x= \frac{\pi}{2}, \frac{3 \pi}{2}, \frac{7 \pi}{6}, \frac{11 \pi}{6}$$
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