Giải phương trình \(\cos x + \cos 3x + 2\cos 5x = 0\).

 \(\begin{array}{l}\,\,\,\,\,\cos x + \cos 3x + 2\cos 5x = 0\\ \Leftrightarrow \cos x + \cos 3x + \cos 5x + \cos 5x = 0\\ \Leftrightarrow \left( {\cos x + \cos 5x} \right) + \left( {\cos 3x + \cos 5x} \right) = 0\\ \Leftrightarrow 2\cos 3x\cos 2x + 2\cos 4x\cos x = 0\\ \Leftrightarrow 2\left( {4{{\cos }^3}x - 3\cos x} \right)\cos 2x + 2\cos 4x\cos x = 0\\ \Leftrightarrow 2\cos x\left( {4{{\cos }^2}x - 3} \right)\cos 2x + 2\cos 4x\cos x = 0\\ \Leftrightarrow 2\cos x\left[ {\left( {4{{\cos }^2}x - 3} \right)\cos 2x + \cos 4x} \right] = 0\\ \Leftrightarrow 2\cos x\left[ {\left[ {2\left( {1 + \cos 2x} \right) - 3} \right]\cos 2x + 2{{\cos }^2}2x - 1} \right] = 0\\ \Leftrightarrow 2\cos x\left[ {\left( {2\cos 2x - 1} \right)\cos 2x + 2{{\cos }^2}2x - 1} \right] = 0\\ \Leftrightarrow 2\cos x\left[ {4{{\cos }^2}2x - \cos 2x - 1} \right] = 0\\ \Leftrightarrow \left[ \begin{array}{l}\cos x = 0\\\cos 2x = \dfrac{{1 + \sqrt {17} }}{8}\\\cos 2x = \dfrac{{1 - \sqrt {17} }}{8}\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = \dfrac{\pi }{2} + k\pi \\2x =  \pm \arccos \dfrac{{1 + \sqrt {17} }}{8} + k2\pi \\2x =  \pm \arccos \dfrac{{1 - \sqrt {17} }}{8} + k2\pi \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}x = \dfrac{\pi }{2} + k\pi \\x =  \pm \dfrac{1}{2}\arccos \dfrac{{1 + \sqrt {17} }}{8} + k\pi \\x =  \pm \dfrac{1}{2}\arccos \dfrac{{1 - \sqrt {17} }}{8} + k\pi \end{array} \right.\end{array}\)

Vậy nghiệm của phương trình là: \(x = \dfrac{\pi }{2} + k\pi \), \(x =  \pm \dfrac{1}{2}\arccos \dfrac{{1 + \sqrt {17} }}{8} + k\pi \), \(x =  \pm \dfrac{1}{2}\arccos \dfrac{{1 - \sqrt {17} }}{8} + k\pi \).