Ta có:
$y=\dfrac{\dfrac{1}{2}}{\dfrac{1}{2}\sin x+\dfrac{\sqrt3}{2}\cos x+2}$
$=\dfrac{\dfrac{1}{2}}{\sin(x+\dfrac{\pi}{3})+2}$
Vì $-1\le \sin(x+\dfrac{\pi}{3})\le 1\Rightarrow \dfrac{1}{6}\le y\le\dfrac{1}{2}$
Vậy $\min y=\dfrac{1}{6} \Leftrightarrow \sin(x+\dfrac{\pi}{3})=1 \Leftrightarrow x=\dfrac{\pi}{6}+k2\pi,k\in\mathbb{Z}$
$\max y=\dfrac{1}{2} \Leftrightarrow \sin(x+\dfrac{\pi}{3})=-1 \Leftrightarrow x=\dfrac{7\pi}{6}+k2\pi,k\in\mathbb{Z}$