Answer
Relative maximum: $x=\frac{\pi}{3}$
Relative minimum: $x=\frac{5 \pi}{3}$
Work Step by Step
First derivative
\[
\frac{-1+2 \cos x}{(2-\cos x)^{2}}=f^{\prime}(x)
\]
Critical points (zeros of the first derivative or the point where the first derivative does not exist):
\[
x=\frac{5 \pi}{3}, x=\frac{\pi}{3}
\]
Second derivative
\[
\frac{2 \sin x(1+\cos x)}{(-2+\cos x)^{3}}=f^{\prime \prime}(x)
\]
Values of second derivative at critical points:
\[
\begin{array}{l}
f^{\prime \prime}\left(\frac{\pi}{3}\right)<0 \\
f^{\prime \prime}\left(\frac{5 \pi}{3}\right)>0
\end{array}
\]
The second derivative tests for points where the second derivative is nonzero, and the first derivative tests for points where the second derivative is zero.
Relative maximum: $x=\frac{\pi}{3}$
Relative minimum: $x=\frac{5 \pi}{3}$
