cos^2x.cos6x-1=0 giải giúp ạ mình đang cần rất gấp

\[\begin{array}{l} {\cos ^2}x.\cos 6x - 1 = 0\\ \Leftrightarrow \frac{{1 + \cos 2x}}{2}\left( {4{{\cos }^3}2x - 3\cos 2x} \right) - 1 = 0\\ \Leftrightarrow \left( {1 + \cos 2x} \right)\left( {4{{\cos }^3}2x - 3\cos 2x} \right) - 2 = 0\\ \Leftrightarrow 4{\cos ^3}2x - 3\cos 2x + 4{\cos ^4}2x - 3{\cos ^2}2x - 2 = 0\\ \Leftrightarrow 4{\cos ^4}2x + 4{\cos ^3}x - 3{\cos ^2}2x - 3\cos 2x - 2 = 0\\ \Leftrightarrow \left( {\cos 2x - 1} \right)\left( {4{{\cos }^3}2x + 8{{\cos }^2}2x + 5\cos 2x + 2} \right) = 0\\ \Leftrightarrow \cos 2x = 1\\ \Leftrightarrow 2x = k2\pi \\ \Leftrightarrow x = k\pi \,\,\,\,\left( {k \in Z} \right). \end{array}\]