Phương trình đã cho tương đương với:
$3(1-\sqrt3)\cos2x+3(1+\sqrt3)(1+\sin2x)=8(\sin x+\cos x)(\sqrt3\sin^3x+\cos^3x)$
$\Leftrightarrow 3(1-\sqrt3)(\cos x-\sin x)(\cos x+\sin x)+3(1+\sqrt3)(\sin x+\cos x)^2=8(\sin x+\cos x)(\sqrt3\sin^3x+\cos^3x)$
$\Leftrightarrow (\sin x+\cos x)[3(1-\sqrt3)(\cos x-\sin x)+3(1+\sqrt3)(\sin x+\cos x)-8(\sqrt3\sin^3x+\cos^3x)]=0$
$\Leftrightarrow (\sin x+\cos x)(6\cos x+6\sqrt3\sin x-8\sqrt3\sin^3x-8\cos^3x)=0$
$\Leftrightarrow (\sin x+\cos x)[6\cos x+6\sqrt3\sin x-2\sqrt3(3\sin x-\sin3x)-2(\cos3x+3\cos x)]=0$
$\Leftrightarrow 2(\sin x+\cos x)(\sqrt3\sin3x-\cos3x)=0$
$\Leftrightarrow 2(\sin x+\cos x)\sin(3x-\dfrac{\pi}{6})=0$
$\Leftrightarrow \left[\begin{array}{l}\sin x+\cos x=0\\\sin(3x-\dfrac{\pi}{6})=0\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}\tan x=-1\\\sin(3x-\dfrac{\pi}{6})=0\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}x=\dfrac{-\pi}{4}+k\pi\\x=\dfrac{\pi}{18}+k\dfrac{\pi}{3}\end{array}\right., k\in\mathbb{Z}$