Đáp án + Giải thích các bước giải:

$2\sin x(1 + \cos 2x) + \sin 2x = 1 + 2\cos x$

$\Leftrightarrow 2 \sin x (1 + 2\cos^2 x - 1) + 2\sin x \cos x = 1 + 2\cos x$

$\Leftrightarrow 2 \sin x (2\cos^2 x + \cos x) - (1 + 2\cos x) = 0$

$\Leftrightarrow 2 \sin x \cos x (2 \cos x + 1) - (1 + 2\cos x) = 0$

$\Leftrightarrow \sin 2x (2 \cos x + 1) - (1 + 2\cos x) = 0$

$\Leftrightarrow (\sin 2x - 1)(2 \cos x + 1) = 0$

$\Leftrightarrow$ \(\left[ \begin{array}{l}\sin 2x = 1\\ \cos x=-\dfrac{1}{2}\end{array} \right.\) 

$\Leftrightarrow \left[ \begin{array}{l}2x = \dfrac{\pi}{2} + k2\pi \\ x = \dfrac{2\pi}{3} + k2\pi \\ x = -\dfrac{2\pi}{3} + k2\pi \end{array} \right. (k \in \mathbb{Z})$

$\Leftrightarrow \left[ \begin{array}{l}x = \dfrac{\pi}{4} + k\pi \\ x = \dfrac{2\pi}{3} + k2\pi \\ x = -\dfrac{2\pi}{3} + k2\pi \end{array} \right. (k \in \mathbb{Z})$.

Vậy $x = \dfrac{\pi}{4} + k\pi$ hoặc $x = \dfrac{2\pi}{3} + k2\pi$ hoặc $x = -\dfrac{2\pi}{3} + k2\pi$ với $k \in \mathbb{Z}$