Sử dụng công thức hạ bậc ta có:

\({\cos ^2}\left( {\frac{x}{2} + \frac{\pi }{6}} \right) = {\cos ^2}\left( {\frac{{3x}}{2} + \frac{\pi }{4}} \right)\)

\( \Leftrightarrow \frac{{1 + \cos \left( {x + \frac{\pi }{3}} \right)}}{2} = \frac{{1 + \cos \left( {3x + \frac{\pi }{2}} \right)}}{2}\)

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

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

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

\( \Leftrightarrow \left[ \begin{array}{l} - 2x = \frac{\pi }{6} + k2\pi \\4x = - \frac{{5\pi }}{6} + k2\pi \end{array} \right.\,\left( {k \in \mathbb{Z}} \right)\)

\( \Leftrightarrow \left[ \begin{array}{l}x = - \frac{\pi }{{12}} + k\pi \\x = - \frac{{5\pi }}{{24}} + k\frac{\pi }{2}\end{array} \right.\,\left( {k \in \mathbb{Z}} \right)\).