Đáp án:
$ \left[\begin{array}{l}x = k\pi\\x = \dfrac{5\pi}{6} + k2\pi\end{array}\right.\quad (k \in \Bbb Z)$
Giải thích các bước giải:
$\begin{array}{l}\sin\left(2x + \dfrac{\pi}{6}\right) + 2\sin x = \dfrac12\\ \Leftrightarrow \cos\dfrac{\pi}{6}.\sin2x + \sin\dfrac{\pi}{6}.\cos2x + 2\sin x =\dfrac12\\ \Leftrightarrow \dfrac{\sqrt3}{2}\sin2x + \dfrac{1}{2}\cos2x + 2\sin x = \dfrac12\\ \Leftrightarrow \sqrt3\sin2x + \cos2x + 4\sin x - 1 = 0\\ \Leftrightarrow 2\sqrt3\sin x\cos x + 1 -2\sin^2x + 4\sin x - 1 =0\\ \Leftrightarrow \sqrt3\sin x\cos x -\sin^2x + 2\sin x =0\\ \Leftrightarrow \sin x(\sqrt3\cos x - \sin x + 2) =0\\ \Leftrightarrow \left[\begin{array}{l}\sin x =0\\\sqrt3\cos x - \sin x = -2\end{array}\right.\\ \Leftrightarrow \left[\begin{array}{l}\sin x =0\\\dfrac{\sqrt3}{2}\cos x - \dfrac{1}{2}\sin x = -1\end{array}\right.\\ \Leftrightarrow \left[\begin{array}{l}\sin x =0\\\cos\left(x + \dfrac{\pi}{6}\right) = -1\end{array}\right.\\ \Leftrightarrow \left[\begin{array}{l}x = k\pi\\x + \dfrac{\pi}{6} = \pi + k2\pi\end{array}\right.\\ \Leftrightarrow \left[\begin{array}{l}x = k\pi\\x = \dfrac{5\pi}{6} + k2\pi\end{array}\right.\quad (k \in \Bbb Z) \end{array}$
