Giải phương trình sau: \({\cos ^2}2x = \frac{1}{4}\).

\({\cos ^2}2x = \frac{1}{4} \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{\cos 2x = \frac{1}{2}}\\{\cos 2x = - \frac{1}{2}}\end{array}} \right.\)

\( \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{2x = \pm k2\pi }\\{2x = \pm \frac{{2\pi }}{3} + k2\pi }\end{array}} \right.\left( {k \in \mathbb{Z}} \right) \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{x = \pm \frac{\pi }{6} + k\pi }\\{x = \pm \frac{\pi }{3} + k\pi }\end{array}} \right.\left( {k \in \mathbb{Z}} \right)\)

Vậy phương trình có 4 họ nghiệm \(\left\{ {\frac{{ \pm \pi }}{3} + k\pi ;\frac{{ \pm \pi }}{6} + k\pi } \right\}\left( {k \in \mathbb{Z}} \right)\).