Đáp án:
$\left[\begin{array}{l}x = k\pi\\x = \dfrac{\pi}{6} + k2\pi\\x = \dfrac{5\pi}{6} + k2\pi\end{array}\right.\quad (k\in\Bbb Z)$
Giải thích các bước giải:
$\sin3x + \cos2x = 1 + 2\sin x\cos2x$
$\Leftrightarrow 3\sin x - 4\sin^3x - 2\sin x\cos2x + \cos2x = 1$
$\Leftrightarrow \sin x[3 - 4\sin^2x - 2(1 - 2\sin^2x)] + \cos2x = 1$
$\Leftrightarrow \sin x +\cos2x = 1$
$\Leftrightarrow \sin + 1 - 2\sin^2x = 1$
$\Leftrightarrow \sin x(1 - 2\sin x)= 0$
$\Leftrightarrow \left[\begin{array}{l}\sin x = 0\\\sin x = \dfrac{1}{2}\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}x = k\pi\\x = \dfrac{\pi}{6} + k2\pi\\x = \dfrac{5\pi}{6} + k2\pi\end{array}\right.\quad (k\in\Bbb Z)$
