Giải phương trình: \({\cos ^2}x + {\cos ^2}2x - {\cos ^2}3x - {\cos ^2}4x = 0\)

\(\begin{array}{l}{\cos ^2}x + {\cos ^2}2x - {\cos ^2}3x - {\cos ^2}4x = 0\\ \Leftrightarrow \frac{{1 + \cos 2x}}{2} + \frac{{1 + \cos 4x}}{2} - \frac{{1 + \cos 6x}}{2} - \frac{{1 + \cos 8x}}{2} = 0\\ \Leftrightarrow 1 + \cos 2x + 1 + \cos 4x - 1 - \cos 6x - 1 - \cos 8x = 0\\ \Leftrightarrow \left( {\cos 2x + \cos 4x} \right) - \left( {\cos 6x + \cos 8x} \right) = 0\\ \Leftrightarrow 2\cos 3x\cos x - 2\cos 7x\cos x = 0\\ \Leftrightarrow 2\cos x\left( {\cos 3x - \cos 7x} \right) = 0\\ \Leftrightarrow 2\cos x.\left[ { - 2\sin 5x\sin \left( { - 2x} \right)} \right] = 0\\ \Leftrightarrow 4\cos x\sin 5x\sin 2x = 0\\ \Leftrightarrow \left[ \begin{array}{l}\cos x = 0\\\sin 5x = 0\\\sin 2x = 0\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = \frac{\pi }{2} + k\pi \\5x = k\pi \\2x = k\pi \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}x = \frac{\pi }{2} + k\pi \\x = \frac{{k\pi }}{5}\\x = \frac{{k\pi }}{2}\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = \frac{{k\pi }}{5}\\x = \frac{{k\pi }}{2}\end{array} \right.,k \in \mathbb{Z}\end{array}\)