Giải phương trình:

${\sin ^2}\left( {x + \frac{\pi }{4}} \right) = {\sin ^2}\left( {2x + \frac{\pi }{2}} \right)$;

${\sin ^2}\left( {x + \frac{\pi }{4}} \right) = {\sin ^2}\left( {2x + \frac{\pi }{2}} \right)$

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

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

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

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

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

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

$\Leftrightarrow x = - \frac{\pi }{4} + k\frac{\pi }{3}\,\,\,\left( {k \in \mathbb{Z}} \right)$.