Đáp án:
$x = k\pi $, $x = \dfrac{\pi }{6} + k2\pi $, $x = \dfrac{{5\pi }}{6} + k2\pi$
Giải thích các bước giải:
$\begin{array}{l} \sin \left( {2x + \dfrac{{9\pi }}{2}} \right) - 3\cos \left( {x - \dfrac{{15\pi }}{2}} \right) = 1 + 2\sin x\\ \Leftrightarrow \cos 2x + 3\sin x = 1 + 2\sin x\\ \Leftrightarrow \left( {1 - 2{{\sin }^2}x} \right) + \sin x - 1 = 0\\ \Leftrightarrow 2{\sin ^2}x - \sin x = 0\\ \Leftrightarrow \left[ \begin{array}{l} \sin x = 0\\ \sin x = \dfrac{1}{2} \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l} x = k\pi \\ x = \dfrac{\pi }{6} + k2\pi \\ x = \dfrac{{5\pi }}{6} + k2\pi \end{array} \right. \end{array}$
Vậy $x = k\pi $, $x = \dfrac{\pi }{6} + k2\pi $, $x = \dfrac{{5\pi }}{6} + k2\pi$.
