$cos\frac{5\pi}{6}=-\frac{\sqrt3}{2}=2cos^2\frac{5\pi}{12}-1=1-2sin^2\frac{5\pi}{12}$$\begin{cases}cos^2\frac{5\pi}{12}=\frac{2-\sqrt3}{4} \\ sin^2\frac{5\pi}{12}=\frac{2+\sqrt3}{4} \end{cases}\Rightarrow \begin{cases}cos\frac{5\pi}{12}=\frac{\sqrt3-1}{2\sqrt2} \\ sin\frac{5\pi}{12}=\frac{\sqrt3+1}{2\sqrt2} \end{cases}$
$2\sqrt2cos(\frac{5\pi}{12}-x)=2\sqrt2cos\frac{5\pi}{12}cosx+2\sqrt2sin\frac{5\pi}{12}sinx$
$=(\sqrt3-1)cosx+(\sqrt3+1)sinx$
Vế trái $=(\sqrt3-1)cosx.sinx+(\sqrt3+1)sin^2x=1$
$\Rightarrow \frac{\sqrt3-1}{2}sin{2x}+\frac{\sqrt3+1}{2}(1-cos2x)=1$
$\Rightarrow \frac{\sqrt3-1}{2}sin2x-\frac{\sqrt3+1}{2}cos2x=\frac{1-\sqrt3}{2}$
$\Rightarrow cos\frac{5\pi}{12}sin2x-sin\frac{5\pi}{12}cos2x=-cos\frac{5\pi}{12}$
$\Rightarrow sin(\frac{5\pi}{12}-2x)=cos\frac{5\pi}{12}=sin\frac{\pi}{12}$
$\Rightarrow \frac{5\pi}{12}-2x=\frac{\pi}{12}+2k\pi$ hoặc $\frac{5\pi}{12}-2x=\frac{11\pi}{12}+2k\pi$
$\Rightarrow x=\frac{\pi}{6}-k\pi , -\frac{\pi}{4}-k\pi$