Giải phương trình: cos2x + cosx = \(4{\sin ^2}\left( {\frac{x}{2}} \right) - 1\).

cos2x + cosx = \(4{\sin ^2}\left( {\frac{x}{2}} \right) - 1\)

⇔ 2cos²x – 1 + cosx = \(2\left[ {2{{\sin }^2}\left( {\frac{x}{2}} \right) - 1} \right] + 1\)

⇔ 2cos²x – 1 + cosx = –2cosx + 1

⇔ 2cos²x + 3cosx – 2 = 0

⇔ (2cosx – 1)(cosx + 2) = 0

⇔ \(\left[ \begin{array}{l}\cos x = \frac{{ - 1}}{2}\\\cos x = - 2\left( L \right)\end{array} \right.\)

⇔ \(x = \pm \frac{\pi }{3} + k2\pi \,\,\left( {k \in \mathbb{Z}} \right)\).