$\frac{\sqrt{3}\sin x}{2+\cos x}$ This is of the form $\frac{u}{v}$
Differentiating above equation and equating to $0$ we get
$\frac{(2+\cos x)\sqrt{3}\cos x - \sqrt{3}\sin x(-\sin x)}{(2+\cos x)^{2}}= 0$
$\implies 2\sqrt{3}\cos x + \sqrt{3}\cos^{2}x + \sqrt{3}\sin^{2}x = 0$
$\implies 2\sqrt{3}\cos x + \sqrt{3} = 0$
$\implies 2\cos x + 1 = 0$
$\implies x = \frac{2π}{3} , \frac{-2π}{3}$
for $x= \frac{2π}{3}$ , $\frac{\sqrt{3}\sin x}{2+\cos x} = 1 $ and
for $x= \frac{-2π}{3}$ , $\frac{\sqrt{3}\sin x}{2+\cos x} = -1$
Hence, option $C$ is correct.