## Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning

# APPENDIX D - Trigonometry - D Exercises: 70

#### Answer

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

#### Work Step by Step

We need to find the range for $x$ for the equation $2cosx+sin2x=0$ $2cosx(sinx+1)=0$ Here, $cosx=0$ gives $x =\frac{ \pi}{2},\frac{3 \pi}{2}$ and $1+sinx=0$ gives $sinx=-1$ Hence, x = $\frac{ \pi}{2},\frac{3 \pi}{2}$

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