Đáp án:
$\begin{array}{l}
3\cos 2x + 2\sin 2x - 3\sin 2x = 2\\
\Rightarrow 3\cos 2x - \sin 2x = 2\\
\Rightarrow \frac{3}{{\sqrt {10} }}\cos 2x - \frac{1}{{\sqrt {10} }}\sin 2x = \frac{2}{{\sqrt {10} }}\\
\Rightarrow \arccos \left( {\frac{3}{{\sqrt {10} }}} \right).\cos 2x - \arcsin \frac{1}{{\sqrt {10} }}.\sin 2x = \frac{2}{{\sqrt {10} }}\\
\Rightarrow \cos \left( {2x + {{18}^0}} \right) = \cos {51^0}\\
\Rightarrow \left[ \begin{array}{l}
2x + {18^0} = {51^0} + k{.360^0}\\
2x + {18^0} = - {51^0} + k{.360^0}
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x = {16^0} + k{.180^0}\\
x = - {35^0} + k{.180^0}
\end{array} \right.
\end{array}$
