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