Answer
Relative maximum: $x=\frac{5 \pi}{6}$
Relative minimum: $x=\frac{7 \pi}{6}$
Work Step by Step
First derivative
\[
2 \cos x+\sqrt{3}=f^{\prime}(x)
\]
Critical points (zeros of first derivative or point where first derivative does not exist)
\[
x=\frac{7 \pi}{6} \text { and } x=\frac{5 \pi}{6}
\]
Second derivative
\[
-2 \sin x=f^{\prime \prime}(x)
\]
Values of second derivative at critical points:
\[
f^{\prime \prime}\left(\frac{5 \pi}{6}\right)=-1<0
\]
\[
f^{\prime \prime}\left(\frac{7 \pi}{6}\right)=1>0
\]
Relative maximum: $x=\frac{5 \pi}{6}$
Relative minimum: $x=\frac{7 \pi}{6}$
