Hướng dẫn giải:

$\begin{align}  & \Leftrightarrow \frac{\sqrt{3}}{2}\sin 4x-\frac{1}{2}\cos 4x=\frac{1}{2}\sin x-\frac{\sqrt{3}}{2}\cos x \\ & \Leftrightarrow \sin (4x-\frac{\pi }{6})=-\cos (x+\frac{\pi }{6}) \\ & \Leftrightarrow \sin (4x-\frac{\pi }{6})=\sin (x-\frac{\pi }{3}) \\ & \Leftrightarrow \left[ \begin{matrix}   4x-\frac{\pi }{6}=x-\frac{\pi }{3}+k2\pi   \\   4x-\frac{\pi }{6}=\frac{4\pi }{3}-x+k2\pi   \\\end{matrix} \right.,k\in \mathbb{Z} \\ & \Leftrightarrow \left[ \begin{matrix}   x=\frac{-\pi }{18}+\frac{k2\pi }{3}  \\   x=\frac{3\pi }{10}+\frac{k2\pi }{5}  \\\end{matrix} \right.,k\in \mathbb{Z} \\\end{align}$

Chọn D.