\({\cos ^2}\left( {2x + \frac{\pi }{2}} \right) = {\sin ^2}\left( {x + \frac{\pi }{6}} \right)\)

\( \Leftrightarrow \frac{{1 + \cos \left( {4x + \pi } \right)}}{2} = \frac{{1 - \cos \left( {2x + \frac{\pi }{3}} \right)}}{2}\) (sử dụng công thức hạ bậc)

\( \Leftrightarrow \cos \left( {4x + \pi } \right) = - \cos \left( {2x + \frac{\pi }{3}} \right)\)

\( \Leftrightarrow \cos \left( {4x + \pi } \right) = \cos \left( {2x + \frac{\pi }{3} + \pi } \right)\) (sử dụng quan hệ hơn kém π)

\( \Leftrightarrow \cos \left( {4x + \pi } \right) = \cos \left( {2x + \frac{{4\pi }}{3}} \right)\)

\( \Leftrightarrow \left[ \begin{array}{l}4x + \pi = 2x + \frac{{4\pi }}{3} + k2\pi \\4x + \pi = - \left( {2x + \frac{{4\pi }}{3}} \right) + k2\pi \end{array} \right.\left( {k \in \mathbb{Z}} \right)\)

\( \Leftrightarrow \left[ \begin{array}{l}4x - 2x = \frac{{4\pi }}{3} - \pi + k2\pi \\4x + 2x = - \frac{{4\pi }}{3} - \pi + k2\pi \end{array} \right.\left( {k \in \mathbb{Z}} \right)\)

\( \Leftrightarrow \left[ \begin{array}{l}2x = \frac{\pi }{3} + k2\pi \\6x = - \frac{{7\pi }}{3} + k2\pi \end{array} \right.\left( {k \in \mathbb{Z}} \right)\)

\( \Leftrightarrow \left[ \begin{array}{l}x = \frac{\pi }{6} + k\pi \\x = - \frac{{7\pi }}{{18}} + k\frac{\pi }{3}\end{array} \right.\left( {k \in \mathbb{Z}} \right)\)

Giải phương trình: căn bậc hai 3 tan x - 1 = 0