Answer
$$ x=\frac{7\pi }{6}+2\pi n,\:x=\frac{11\pi }{6}+2\pi n $$
Work Step by Step
Given
$$ y=\frac{\cos x}{2+\sin x} $$
Since
\begin{aligned}
\frac{dy}{dx}&= \frac{\frac{d}{dx}\left(\cos \left(x\right)\right)\left(2+\sin \left(x\right)\right)-\frac{d}{dx}\left(2+\sin \left(x\right)\right)\cos \left(x\right)}{\left(2+\sin \left(x\right)\right)^2}\\
&= \frac{\left(-\sin \left(x\right)\right)\left(2+\sin \left(x\right)\right)-\cos \left(x\right)\cos \left(x\right)}{\left(2+\sin \left(x\right)\right)^2}\\
&= \frac{-\sin \left(x\right)\left(2+\sin \left(x\right)\right)-\cos ^2\left(x\right)}{\left(2+\sin \left(x\right)\right)^2}
\end{aligned}
The tangent is horizontal when $y'=0 $, then
\begin{aligned}
y'&=0\\
-\sin(x)(2+\sin x)-\cos^2(x)&=0 \\
-2\sin(x)-\sin^2(x)-(1-\sin^2(x))&=0\\
-2\sin(x)-1&=0\\
\sin(x)&=\frac{-1}{2}
\end{aligned}
It follows that
$$ x=\frac{7\pi }{6}\ \ \ \ \ ,\:x=\frac{11\pi }{6}$$
and the general
$$ x=\frac{7\pi }{6}+2\pi n,\:x=\frac{11\pi }{6}+2\pi n $$
