ta có :$2\sqrt{2}(\frac{1}{2}\sin x+\frac{\sqrt{3}}{2}\cos x)=2(\frac{\sqrt{3}}{2}\cos 2x-\frac{1}{2}\sin 2x$
$\Leftrightarrow 2\sqrt{2}(\cos \frac{pi}{3}\sin x+\sin \frac{pi}{3}\cos x)=2(\sin \frac{pi}{3}\cos 2x-\cos \frac{pi}{3}\sin 2x)$
$\Leftrightarrow 2\sqrt{2}(\sin (\frac{pi}{3}+x)=2\sin (\frac{pi}{3}-2x)$
$\Leftrightarrow \sqrt{2}\cos (\frac{pi}{2}-\frac{pi}{3}-x)=\sin (2(\frac{pi}{6}-x))$
$\Leftrightarrow \sqrt{2}\cos (\frac{pi}{6}-x)=2sin(\frac{pi}{6}-x)\cos (\frac{pi}{6}-x)$
TH1:
$\cos(\frac{pi}{6}-x)=0$
$\Leftrightarrow x=-\frac{pi}{3}-kpi$
TH2:
$\sin (\frac{pi}{6}-x)=\frac{1}{\sqrt{2}}$
$\Leftrightarrow \sin (\frac{pi}{6}-x)=sin\frac{pi}{4}$
$\Leftrightarrow x=-\frac{1}{12}-k2pi hoặc x=\frac{-7}{12}-k2pi$