Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
g,\\
{\sin ^2}2x + {\cos ^2}3x = 1\\
\Leftrightarrow {\sin ^2}2x + \left( {1 - {{\sin }^2}3x} \right) = 1\\
\Leftrightarrow {\sin ^2}2x - {\sin ^2}3x = 0\\
\Leftrightarrow \left( {\sin 2x - \sin 3x} \right)\left( {\sin 2x + \sin 3x} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\sin 2x = \sin 3x\\
\sin 2x = - \sin 3x
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
\sin 2x = \sin 3x\\
\sin 2x = \sin \left( { - 3x} \right)
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
2x = 3x + k2\pi \\
2x = \pi - 3x + k2\pi \\
2x = - 3x + k2\pi \\
2x = \pi + 3x + k2\pi
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = k2\pi \\
x = \dfrac{\pi }{5} + \dfrac{{k2\pi }}{5}\\
x = \dfrac{{k2\pi }}{5}\\
x = - \pi + k2\pi
\end{array} \right. \Leftrightarrow x = \dfrac{{k\pi }}{5}\\
i,\\
5\cos 2x - 12\sin 2x = - 13\\
\Leftrightarrow \dfrac{5}{{13}}\cos 2x - \dfrac{{12}}{{13}}\sin 2x = - 1\\
\sin t = \dfrac{5}{{13}};\,\,\,\,\cos t = \dfrac{{12}}{{13}}\,\,\,\,\,\,\,\,\,\,\left( {{{\sin }^2}t + {{\cos }^2}t = 1} \right)\\
\Rightarrow \sin t.\cos 2x - \cos t.\sin 2x = - 1\\
\Leftrightarrow \sin \left( {t - 2x} \right) = \sin \dfrac{{ - \pi }}{2}\\
\Leftrightarrow t - 2x = - \dfrac{\pi }{2} + k2\pi \\
\Leftrightarrow 2x = t + \dfrac{\pi }{2} + k2\pi \\
\Leftrightarrow x = \dfrac{t}{2} + \dfrac{\pi }{4} + k\pi \\
\Leftrightarrow x = \dfrac{{\arcsin \dfrac{5}{{13}}}}{2} + \dfrac{\pi }{4} + k\pi \\
k,\\
2\cos x + 3\sin x = 2\\
\Leftrightarrow \cos x = \dfrac{{2 - 3\sin x}}{2}\\
{\sin ^2}x + {\cos ^2}x = 1\\
\Leftrightarrow {\sin ^2}x + {\left( {\dfrac{{2 - 3\sin x}}{2}} \right)^2} = 1\\
\Leftrightarrow {\sin ^2}x + {\left( {1 - \dfrac{3}{2}\sin x} \right)^2} = 1\\
\Leftrightarrow {\sin ^2}x + 1 - 3\sin x + \dfrac{9}{4}{\sin ^2}x = 1\\
\Leftrightarrow \dfrac{{13}}{4}{\sin ^2}x - 3\sin x = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\sin x = 0\\
\sin x = \dfrac{{12}}{{13}}\,\,\,\,\,\left( {\cos x = - \dfrac{5}{{13}}} \right)
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = k\pi \\
x = \pi - \arcsin \dfrac{{12}}{{13}} + k2\pi
\end{array} \right.
\end{array}\)
