Giải phương trình: cos4x + cos2x + 1 = 0.

cos4x + cos2x + 1 = 0 \( \Leftrightarrow 2{\cos ^2}2x - 1 + \cos 2x + 1 = 0\)

\( \Leftrightarrow 2{\cos ^2}2x + \cos 2x = 0 \Leftrightarrow \cos 2x\left( {2\cos 2x + 1} \right) = 0\)

\( \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{\cos 2x = 0\left( 1 \right)}\\{2\cos 2x + 1 = 0\left( 2 \right)}\end{array}} \right.\)

\(\left( 1 \right) \Leftrightarrow 2x = \frac{\pi }{2} + k\pi \left( {k \in \mathbb{Z}} \right) \Leftrightarrow x = \frac{\pi }{4} + k\frac{\pi }{2}\left( {k \in \mathbb{Z}} \right)\)

\(\left( 2 \right) \Leftrightarrow \cos 2x = - \frac{1}{2} = \cos \frac{{2\pi }}{3}\)\( \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{2x = \frac{{2\pi }}{3} + k2\pi }\\{2x = - \frac{{2\pi }}{3} + k2\pi }\end{array}} \right.\left( {k \in \mathbb{Z}} \right) \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{x = \frac{\pi }{3} + k\pi }\\{x = - \frac{\pi }{3} + k\pi }\end{array}} \right.\left( {k \in \mathbb{Z}} \right)\).