## Calculus: Early Transcendentals 8th Edition

$x=\frac{2\pi}{3}+2\pi k, \frac{4\pi}{3}+2\pi k$
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 $f'(x)=1+2\cos{x}$. $f'(x)=1+2\cos{x}=0 \Rightarrow \cos{x}=-\frac{1}{2} \Rightarrow x=\frac{2\pi}{3}+2\pi k, \frac{4\pi}{3}+2\pi k$.