ĐK $\sin 2x \ne 0 \Leftrightarrow x \ne \dfrac{k\pi}{2}$
$8\sin^2 x \cos x =\sqrt 3 \sin x +\cos x$
$\Leftrightarrow 4\sin x \sin 2x =\sqrt 3 \sin x + \cos x$
$\Leftrightarrow 2(\cos x -\cos 3x) = \sqrt 3 \sin x + \cos x$
$\Leftrightarrow \cos x + \sqrt 3 \sin x = 2\cos 3x$
$\Leftrightarrow \dfrac{1}{2}\cos x +\dfrac{\sqrt 3 }{2} \sin x = \cos 3x$
$\Leftrightarrow \cos (x-\dfrac{\pi}{3})=\cos 3x$ tự làm nốt đi