Giải phương trình: sin 3x + 2 sin^2 x − 1 = 0.
Giải thích
\[\sin 3x + 2{\sin ^2}x - 1 = 0 \Leftrightarrow \sin 3x + 2\frac{{1 - \cos 2x}}{2} - 1 = 0 \Leftrightarrow \sin 3x = \cos 2x\]
\[ \Leftrightarrow \sin 3x = \sin \left( {\frac{\pi }{2} - 2x} \right) \Leftrightarrow \left[ \begin{array}{l}3x = \frac{\pi }{2} - 2x + k2\pi \\3x = \pi - \frac{\pi }{2} + 2x + k2\pi \end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = \frac{\pi }{{10}} + k\frac{{2\pi }}{5}\\x = \frac{\pi }{2} + k2\pi \end{array} \right.,\,\,k \in \mathbb{Z}\]
\[ \Rightarrow x = \frac{\pi }{{10}} + k\frac{{2\pi }}{5},\,\,\,\,k \in \mathbb{Z}\]