$2-\sqrt3cos2x+sin2x=4cos^23x$$\Leftrightarrow 1-(\frac{\sqrt3}{2}cos2x-\frac{1}{2}sin2x)=2cos^23x$
$\Leftrightarrow -(cos\frac{\pi}{6}cos2x-sin\frac{\pi}{6}sin2x)=2cos^23x-1$
$\Leftrightarrow -cos(2x+\frac{\pi}{6})=cos6x$
$\Leftrightarrow cos(2x-\frac{5\pi}{6})=cos6x$
$\Leftrightarrow 2x-\frac{5\pi}{6}+2k\pi=6x $ hoặc $2x-\frac{5\pi}{6}+6x=2k\pi$
$\Leftrightarrow x=\frac{k\pi}{2}-\frac{5\pi}{24}$ hoặc $x=\frac{k\pi}{4}+\frac{5\pi}{48}$