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.$$\Leftrightarrow \left[\begin{array}{l}x=\dfrac{-\pi}{4}+k\pi\\x=\dfrac{\pi}{18}+k\dfrac{\pi}{3}\end{array}\right., k\in\mathbb{Z}$
Phương trình đã cho tương đương với: $3(1-\sqrt3)\cos2x+3(1+\sqrt3)(1+\sin2x)=8(\sin x+\cos x)(\sqrt3\sin^3x+\cos^3x)$$\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\sqrt3(\sin x+\cos x)(\sin3x-\cos3x)=0$$\Leftrightarrow \left[\begin{array}{l}\sin x+\cos x=0\\\sin3x=\cos3x\end{array}\right.$$\Leftrightarrow \left[\begin{array}{l}\tan x=-1\\\tan3x=1\end{array}\right.$$\Leftrightarrow \left[\begin{array}{l}x=\dfrac{-\pi}{4}+k\pi\\x=\dfrac{\pi}{12}+k\dfrac{\pi}{3}\end{array}\right., k\in\mathbb{Z}$
Phương trình đã cho tương đương với: $3(1-\sqrt3)\cos2x+3(1+\sqrt3)(1+\sin2x)=8(\sin x+\cos x)(\sqrt3\sin^3x+\cos^3x)$$\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}{1
8}+k\dfrac{\pi}{3}\end{array}\right., k\in\mathbb{Z}$