Đáp án:
$\left[\begin{array}{l}x = -\dfrac{\pi}{4} + k\pi\\x = -\dfrac{\pi}{6} + k2\pi\\x = \dfrac{7\pi}{6} + k2\pi\end{array}\right.\quad (k\in\Bbb Z)$
Giải thích các bước giải:
$2\sin^2x + \sin2x + \sin x+\cos x = 0$
$\Leftrightarrow 2\sin^2x + 2\sin x\cos x + \sin x+\cos x = 0$
$\Leftrightarrow 2\sin x(\sin x + \cos x) + (\sin x+\cos x)= 0$
$\Leftrightarrow (\sin x+\cos x)(2\sin x + 1) = 0$
$\Leftrightarrow \left[\begin{array}{l}\sin x +\cos x = 0\\2\sin x +1 = 0\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}\sin\left(x +\dfrac{\pi}{4}\right)= 0\\\sin x = -\dfrac{1}{2}\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}x = -\dfrac{\pi}{4} + k\pi\\x = -\dfrac{\pi}{6} + k2\pi\\x = \dfrac{7\pi}{6} + k2\pi\end{array}\right.\quad (k\in\Bbb Z)$
