Đáp án:
$\begin{array}{l}
g){\sin ^2}2x + {\cos ^2}3x = 1\\
\Rightarrow {\sin ^2}2x + {\cos ^2}3x = {\sin ^2}2x + {\cos ^2}2x\\
\Rightarrow {\cos ^2}3x = {\cos ^2}2x\\
\Rightarrow \left[ \begin{array}{l}
\cos 3x = \cos 2x\\
\cos 3x = - \cos 2x \Rightarrow cos3x = cos\left( {2x + \pi } \right)
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
3x = 2x + k2\pi \\
3x = - 2x + k2\pi \\
3x = 2x + \pi + k2\pi \\
3x = - 2x - \pi + k2\pi
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x = k2\pi \\
x = \dfrac{{k2\pi }}{5}\\
x = \pi + k2\pi \\
x = - \dfrac{\pi }{5} + \dfrac{{k2\pi }}{5}
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x = \dfrac{{k2\pi }}{5}\\
x = - \dfrac{\pi }{5} + \dfrac{{k2\pi }}{5}
\end{array} \right.\\
i)5\cos 2x - 12\sin 2x = - 13\\
\Rightarrow \dfrac{5}{{12}}.\cos 2x - \dfrac{{12}}{{13}}.\sin 2x = - 1\\
\Rightarrow \cos \left( {\arccos \dfrac{5}{{12}}} \right).\cos 2x - \sin \left( {\arccos \dfrac{5}{{12}}} \right).\sin 2x = - 1\\
\Rightarrow \cos \left( {2x + \arccos \dfrac{5}{{12}}} \right) = - 1\\
\Rightarrow 2x + \arccos \dfrac{5}{{12}} = \pi + k2\pi \\
\Rightarrow x = - \dfrac{1}{2}.\arccos \dfrac{5}{{12}} + \dfrac{\pi }{2} + k\pi
\end{array}$
