$\begin{array}{l} 2{\cos ^2}\left( {\dfrac{\pi }{4} - 2x} \right) + \sqrt 3 {\cos}4x = 4{\cos ^2}x - 1\\ \Leftrightarrow \left( {1 + \cos \left( {\dfrac{\pi }{2} - 4x} \right)} \right) + \sqrt 3 {\cos ^4}x = 4{\cos ^2}x - 1\\ \Leftrightarrow 1 + \sin 4x + \sqrt 3 \cos 4x = 4{\cos ^2}x - 1\\ \Leftrightarrow 1 + \sin 4x + \sqrt 3 .\cos 4x = \cos 2x + 2{\cos ^2}x\\ \Leftrightarrow 1 + \sin 4x + \sqrt 3 \cos 4x = \cos 2x + 2{\cos ^2}x\\ \Leftrightarrow 1 - 2{\cos ^2}x + \sin 4x + \sqrt 3 \cos 4x = \cos 2x\\ \Leftrightarrow - \cos 2x + 2\sin 2x\cos 2x + = \cos 2x\\ \Leftrightarrow 2\cos 2x = \sin 4x + \sqrt 3 \cos 4x\\ \Leftrightarrow 2\cos 2x = 2\cos \left( {4x - \dfrac{\pi }{6}} \right)\\ \Leftrightarrow \left[ \begin{array}{l} 4x - \dfrac{\pi }{6} = 2x + k2\pi \\ 4x - \dfrac{\pi }{6} = - 2x + k2\pi \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l} x = \dfrac{\pi }{{12}} + k\pi \\ x = \dfrac{\pi }{{36}} + \dfrac{{k\pi }}{3} \end{array} \right.\left( {k \in \mathbb{Z}} \right) \end{array}$