$\begin{array}{l}
8{\cos ^3}\left( {x - \dfrac{\pi }{3}} \right) - 3\cos x - 3\sqrt 3 \sin x + 2\cos 2x = 0\\
\Leftrightarrow 8{\cos ^3}\left( {x - \dfrac{\pi }{3}} \right) - 3\left( {\cos x + \sqrt 3 \sin x} \right) + 2\cos 2x = 0\\
\Leftrightarrow 8{\cos ^3}\left( {x - \dfrac{\pi }{3}} \right) - 6.\left( {\dfrac{1}{2}\cos x + \dfrac{{\sqrt 3 }}{2}\sin x} \right) + 2\cos 2x = 0\\
\Leftrightarrow 8{\cos ^3}\left( {x - \dfrac{\pi }{3}} \right) - 6.\cos \left( {x - \dfrac{\pi }{3}} \right) + 2\cos 2x = 0\\
\Leftrightarrow 2\left[ {4{{\cos }^3}\left( {x - \dfrac{\pi }{3}} \right) - 3\cos \left( {x - \dfrac{\pi }{3}} \right)} \right] + 2\cos 2x = 0\\
\Leftrightarrow 2.\cos \left( {3x - 3.\dfrac{\pi }{3}} \right) + 2\cos 2x = 0\\
\Leftrightarrow 2\cos \left( {3x - \pi } \right) + 2\cos 2x = 0\\
\Leftrightarrow - \cos 3x + \cos 2x = 0 \Leftrightarrow \cos 3x = \cos 2x\\
\Leftrightarrow \left[ \begin{array}{l}
3x = 2x + k2\pi \\
3x = - 2x + k2\pi
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = k2\pi \\
x = \dfrac{{k2\pi }}{5}
\end{array} \right. \Leftrightarrow x = \dfrac{{k2\pi }}{5}
\end{array}$
