Giải phương trình: cos3x.cosx = cos2x.

cos3x.cosx = cos2x \( \Leftrightarrow \frac{1}{2}\left( {\cos 4x + \cos 2x} \right) = \cos 2x\)

\( \Leftrightarrow \cos 4x = \cos 2x \Leftrightarrow 2{\cos ^2}2x - 1 - \cos 2x = 0\)

\( \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{\cos 2x = 1}\\{\cos 2x = - \frac{1}{2}}\end{array}} \right.\)

Với cos2x = 1 \( \Rightarrow x = k\pi \,\left( {k \in \mathbb{Z}} \right)\).

Với cos2x = \( - \frac{1}{2} \Rightarrow \left[ {\begin{array}{*{20}{c}}{x = \frac{\pi }{3} + k\pi }\\{x = - \frac{\pi }{3} + k\pi }\end{array}} \right.\left( {k \in \mathbb{Z}} \right)\).