Giải phương trình sau: 2cosx.cos2x = 1 + cos2x + cos3x.

2cosx.cos2x = 1 + cos2x + cos3x

⇔ 2cosx.cos2x = 2cos²x + 4cos³x – 3cosx

⇔ cosx[2(2cos²x – 1) – 2cosx – 4cos²x + 3] = 0

⇔ cosx(–2cosx + 1) = 0

⇔ $\left[ \begin{array}{l}\cos x = 0\\ - 2\cos x = - 1\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = \frac{\pi }{2} + k\pi \\x = \pm \frac{\pi }{3} + k2\pi \end{array} \right.\left( {k \in \mathbb{Z}} \right)$.