Đáp án:
$\left[\begin{array}{l}x = \dfrac{\pi}{2} + k2\pi\\x = \dfrac{\pi}{3} + k2\pi\\x = -\dfrac{\pi}{3} + k2\pi\end{array}\right.\quad (k \in \Bbb Z)$
Giải thích các bước giải:
$1 + 2\sin x\cos x = \sin x + 2\cos x$
$\Leftrightarrow 1 - \sin x + 2\sin x\cos x - 2\cos x = 0$
$\Leftrightarrow (1-\sin x)(1 - 2\cos x) = 0$
$\Leftrightarrow \left[\begin{array}{l}\sin x = 1\\\cos x = \dfrac{1}{2}\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}x = \dfrac{\pi}{2} + k2\pi\\x = \dfrac{\pi}{3} + k2\pi\\x = -\dfrac{\pi}{3} + k2\pi\end{array}\right.\quad (k \in \Bbb Z)$
