$\begin{array}{l} ĐK:\cos x \ne 0 \Leftrightarrow x \ne \dfrac{\pi }{2} + k\pi \\ \left( {2{{\sin }^2}x - 1} \right){\tan ^2}2x + 3\left( {2{{\cos }^2}x - 1} \right) = 0\\ \Leftrightarrow - \left( {1 - 2{{\sin }^2}x} \right){\tan ^2}2x + 3\left( {2{{\cos }^2}x - 1} \right) = 0\\ \Leftrightarrow - \cos 2x.{\tan ^2}2x + 3\cos 2x = 0\\ \Leftrightarrow \cos 2x\left( { - {{\tan }^2}2x + 3} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l} \cos 2x = 0\\ {\tan ^2}2x = 3 \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} 2x = \dfrac{\pi }{2} + k\pi \\ \tan 2x = \sqrt 3 \\ \tan 2x = - \sqrt 3 \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l} x = \dfrac{\pi }{4} + k\dfrac{\pi }{2}\\ 2x = \dfrac{\pi }{3} + k\pi \\ 2x = - \dfrac{\pi }{3} + k\pi \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x = \dfrac{\pi }{4} + k\dfrac{\pi }{2}\\ x = \dfrac{\pi }{6} + \dfrac{{k\pi }}{2}\\ x = - \dfrac{\pi }{6} + \dfrac{{k\pi }}{2} \end{array} \right.\left( {k \in Z} \right) \end{array}$
