$\begin{array}{l} \cos 2x - 2{\cos ^6}x + {\sin ^4}x = 0\\ \Leftrightarrow 2{\cos ^2}x - 1 - 2{\cos ^6}x + {\left( {{{\sin }^2}x + {{\cos }^2}x} \right)^2} - 2{\sin ^2}x{\cos ^2}x - {\cos ^2}x = 0\\ \Leftrightarrow 2{\cos ^2}x\left( {1 - {{\cos }^4}x} \right) - 1 + {\left( {{{\sin }^2}x + {{\cos }^2}x} \right)^2} - 2{\sin ^2}x{\cos ^2}x - {\cos ^2}x = 0\\ \Leftrightarrow 2{\cos ^2}x\left( {1 - {{\cos }^2}x} \right)\left( {1 + {{\cos }^2}x} \right) - 1 + 1 - 2{\sin ^2}x{\cos ^2}x - {\cos ^2}x = 0\\ \Leftrightarrow 2{\cos ^2}x{\sin ^2}x\left( {1 + {{\cos }^2}x} \right) - 2{\cos ^2}x\left( {{{\sin }^2}x - 1} \right) = 0\\ \Leftrightarrow 2{\cos ^2}x{\sin ^2}x\left( {1 + {{\cos }^2}x} \right) - 2{\cos ^2}x\left( { - {{\cos }^2}x} \right) = 0\\ \Leftrightarrow 2{\cos ^2}x{\sin ^2}x\left( {1 + {{\cos }^2}x} \right) + 2{\cos ^4}x = 0\\ \Leftrightarrow 2{\cos ^2}x\left[ {{{\sin }^2}x + {{\sin }^2}x{{\cos }^2}x + {{\cos }^2}x} \right] = 0\\ \Leftrightarrow 2{\cos ^2}x\left( {1 + {{\sin }^2}x{{\cos }^2}x} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l} \cos x = 0\\ 1 + {\sin ^2}x{\cos ^2}x = 0(L)\left( {1 + {{\sin }^2}x{{\cos }^2}x \ge 1 > 0} \right) \end{array} \right.\\ \Leftrightarrow \cos x = 0 \Leftrightarrow x = \dfrac{\pi }{2} + k\pi \left( {k \in \mathbb{Z}} \right) \end{array}$
