`~rai~`

\(y=\sqrt{3}\sin2x-\cos2x\\\quad=2\left(\dfrac{\sqrt{3}}{2}\sin2x-\dfrac{1}{2}\cos2 x\right)\\\quad=2\left(\cos\dfrac{\pi}{6}\sin2x-\sin\dfrac{\pi}{6}\cos2x\right)\\\quad=2\sin\left(2x-\dfrac{\pi}{6}\right)\\\text{Ta có:}-1\le\sin\left(2x-\dfrac{\pi}{6}\right)\le 1\\\Leftrightarrow -2\le2\sin\left(2x-\dfrac{\pi}{6}\right)\le 2\\\Leftrightarrow -2\le y\le 2.\\+)Min_y=-2\Leftrightarrow \sin\left(2x-\dfrac{\pi}{6}\right)=-1\\\Leftrightarrow 2x-\dfrac{\pi}{6}=-\dfrac{\pi}{2}+k2\pi\\\Leftrightarrow 2x=-\dfrac{\pi}{3}+k2\pi\\\Leftrightarrow x=-\dfrac{\pi}{6}+k\pi.(k\in\mathbb{Z})\\+)Max_y=2\Leftrightarrow \sin\left(2x-\dfrac{\pi}{6}\right)=1\\\Leftrightarrow 2x-\dfrac{\pi}{6}=\dfrac{\pi}{2}+k2\pi\\\Leftrightarrow 2x=\dfrac{2\pi}{3}+k2\pi\\\Leftrightarrow x=\dfrac{\pi}{3}+k\pi.\\\text{Vậy Min}_y=-2\text{ khi x=}-\dfrac{\pi}{6}+k\pi;\\\text{Max}_y=2\text{ khi }x=\dfrac{\pi}{3}+k\pi.(k\in\mathbb{Z})\)

$y=\sqrt3\sin2x-\cos2x$

$=2\left(\dfrac{\sqrt3}{2}\sin2x-\dfrac{1}{2}\cos2x\right)$

$=2\sin\left(2x-\dfrac{\pi}{6}\right)$

Ta có $-1\le \sin\left(2x-\dfrac{\pi}{6}\right)\le 1$

$\to -2\le y\le 2$

Vậy:

$\begin{cases}\min y=-2\to \sin\left(2x-\dfrac{\pi}{6}\right)=-1\to x=\dfrac{-\pi}{6}+k\pi \\ \max y=2\to \sin\left(2x-\dfrac{\pi}{6}\right)=1\to x=\dfrac{\pi}{3}+k\pi\end{cases}$