Đáp án:
$\left[\begin{array}{l}x = -\dfrac{\pi}{6} + k2\pi\\x = \dfrac{7\pi}{6} + k2\pi\end{array}\right.\quad (k \in \Bbb Z)$
Giải thích các bước giải:
$2\sin x\cos x + \cos x = 4\sin x +2$
$\Leftrightarrow \cos x(2\sin x +1)= 2(2\sin x +1)$
$\Leftrightarrow (2\sin x +1)(\cos x -2) = 0$
$\Leftrightarrow \left[\begin{array}{l}\sin x = -\dfrac{1}{2}\quad (nhận)\\\cos x= 2\quad (loại)\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}x = -\dfrac{\pi}{6} + k2\pi\\x = \dfrac{7\pi}{6} + k2\pi\end{array}\right.\quad (k \in \Bbb Z)$
