1 câu trả lời
\[\begin{array}{l} 2\sin 2x + 3\cos 2x - 1 = 0\\ \Leftrightarrow 2\sin 2x + 3\cos 2x = 1\\ \Leftrightarrow \frac{2}{{\sqrt {13} }}\sin 2x + \frac{3}{{\sqrt {13} }}\cos 2x = \frac{1}{{\sqrt {13} }}\\ \Leftrightarrow \sin \left( {2x + \alpha } \right) = \frac{1}{{\sqrt {13} }}\,\,\,\,\,\left( {\cos \alpha = \frac{2}{{\sqrt {13} }};\,\,\,\sin \alpha = \frac{3}{{\sqrt {13} }}} \right)\\ \Leftrightarrow \sin \left( {2x + \alpha } \right) = \sin \beta \\ \Leftrightarrow \left[ \begin{array}{l} 2x + \alpha = \beta + k2\pi \\ 2x + \alpha = \pi - \beta + k2\pi \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x = \frac{{\beta - \alpha }}{2} + k\pi \\ x = \frac{{\pi - \beta - \alpha }}{2} + k\pi \end{array} \right.\,\,\,\left( {k \in Z} \right). \end{array}\]
