Giải thích các bước giải:
$\begin{array}{l}
a)\sin \left( {2x + \dfrac{{5\pi }}{2}} \right) - 3\cos \left( {x - \dfrac{{7\pi }}{2}} \right) = 1 + 2\sin x\\
\Leftrightarrow \sin \left( {2x + \dfrac{\pi }{2}} \right) - 3\cos \left( {x + \dfrac{\pi }{2}} \right) = 1 + 2\sin x\\
\Leftrightarrow \cos 2x - 3\sin \left( { - x} \right) = 1 + 2\sin x\\
\Leftrightarrow 1 - 2{\sin ^2}x + \sin x = 1\\
\Leftrightarrow 2{\sin ^2}x - \sin x = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\sin x = 0\\
\sin x = \dfrac{1}{2}
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = k\pi \\
x = \dfrac{\pi }{6} + k2\pi \\
x = \dfrac{{5\pi }}{6} + k2\pi
\end{array} \right.\left( {k \in Z} \right)\\
b)\sin \left( {x + \dfrac{\pi }{6}} \right) - \cos \left( {2x + \dfrac{\pi }{3}} \right) = 2\\
\Leftrightarrow \sin \left( {x + \dfrac{\pi }{6}} \right) - \left( {1 - 2{{\sin }^2}\left( {x + \dfrac{\pi }{6}} \right)} \right) = 2\\
\Leftrightarrow 2{\sin ^2}\left( {x + \dfrac{\pi }{6}} \right) + \sin \left( {x + \dfrac{\pi }{6}} \right) - 3 = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\sin \left( {x + \dfrac{\pi }{6}} \right) = 1\\
\sin \left( {x + \dfrac{\pi }{6}} \right) = \dfrac{{ - 3}}{2}\left( l \right)
\end{array} \right.\\
\Leftrightarrow x + \dfrac{\pi }{6} = \dfrac{\pi }{2} + k2\pi \left( {k \in Z} \right)\\
\Leftrightarrow x = \dfrac{\pi }{3} + k2\pi \left( {k \in Z} \right)
\end{array}$
Câu c bạn xem lại đề.
