Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 3 - Section 3.3 - Derivatives of Trigonometric Functions - 3.3 Exercises - Page 196: 33


$x=\frac{2\pi}{3}+2\pi k, \frac{4\pi}{3}+2\pi k$

Work Step by Step

First, note that to have a horizontal tangent, we must have a slope of zero. Since the derivative of a function gives us the slope at that point we will determine when $f'(x)=0$. If $f(x)=x+2\sin{x}$ then $f'(x)=1+2\cos{x}$. $f'(x)=1+2\cos{x}=0 \Rightarrow \cos{x}=-\frac{1}{2} \Rightarrow x=\frac{2\pi}{3}+2\pi k, \frac{4\pi}{3}+2\pi k$.
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