Giải phương trình: \(2{\sin ^2}\frac{x}{2} = \cos 5x + 1\).

\(2{\sin ^2}\frac{x}{2} = \cos 5x + 1\)

⇔ \(2.\frac{{1 - \cos x}}{2} = \cos 5x + 1\)

⇔ 1 – cosx = cos5x + 1

⇔ cos5x + cosx = 0

⇔ 2cos3x cos2x = 0

⇔ \(\left[ \begin{array}{l}\cos 3x = 0\\\cos 2x = 0\end{array} \right.\)

⇔ \(\left[ \begin{array}{l}3x = \frac{\pi }{2} + k\pi \\2x = \frac{\pi }{2} + k\pi \end{array} \right.\,\,\left( {k \in \mathbb{Z}} \right)\)

⇔ \(\left[ \begin{array}{l}x = \frac{\pi }{6} + k\frac{\pi }{3}\\x = \frac{\pi }{4} + k\frac{\pi }{2}\end{array} \right.\,\left( {k \in \mathbb{Z}} \right)\).