$2\sqrt{3}\sin (x-\dfrac{\pi}{8}).\cos (x-\dfrac{\pi}{8})+2\cos ^2(x-\dfrac{\pi}{8})=\sqrt{3}+4(\sin ^2x+\cos (x+\dfrac{\pi}{3})\sin (x+\dfrac{\pi}{6})) \\
\leftrightarrow \sqrt{3}.\sin (2x-\dfrac{\pi}{4})+1+\cos (2x-\dfrac{\pi}{4})=\sqrt{3}+2(1-\cos 2x)+2(\sin (2x+\dfrac{\pi}{2}-\sin (\dfrac{\pi}{6}))) \\$
$\text{đến đây bạn có thể tự giải nhá}$