$y=\sqrt3\sin2x-\cos2x$
$=2\left(\dfrac{\sqrt3}{2}\sin2x-\dfrac{1}{2}\cos2x\right)$
$=2\sin\left(2x-\dfrac{\pi}{6}\right)$
Ta có $-1\le \sin\left(2x-\dfrac{\pi}{6}\right)\le 1$
$\to -2\le y\le 2$
Vậy:
$\begin{cases}\min y=-2\to \sin\left(2x-\dfrac{\pi}{6}\right)=-1\to x=\dfrac{-\pi}{6}+k\pi \\ \max y=2\to \sin\left(2x-\dfrac{\pi}{6}\right)=1\to x=\dfrac{\pi}{3}+k\pi\end{cases}$