Giải phương trình sau: sin 2x = sin pi/4 .

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

\[ \Leftrightarrow \left[ \begin{array}{l}2x = \frac{\pi }{4} + k2\pi \\2x = \frac{{3\pi }}{4} + k2\pi \end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = \frac{\pi }{8} + k\pi \\x = \frac{{3\pi }}{8} + k\pi \end{array} \right.(k \in \mathbb{Z})\]