Đáp án:
$\left[\begin{array}{l}x = \dfrac{\pi}{4}\\x = \dfrac{3\pi}{4}\\x = \dfrac{\pi}{6}\\x =\dfrac{5\pi}{6}\end{array}\right.$
Giải thích các bước giải:
$\sin3x + \cos2x - \sin x = 0$
$\Leftrightarrow 2\cos2x\sin x - \cos2x = 0$
$\Leftrightarrow \cos2x(2\sin x - 1)=0$
$\Leftrightarrow \left[\begin{array}{l}\cos2x = 0\\\sin x =\dfrac{1}{2}\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}x= \dfrac{\pi}{4} + k\dfrac{\pi}{2}\\x =\dfrac{\pi}{6}+ k2\pi\\x =\dfrac{5\pi}{6} + k2\pi\end{array}\right.\quad (k \in\Bbb Z)$
Ta có:
$0 < x < \pi$
$\to \left[\begin{array}{l}0 < \dfrac{\pi}{4} + k\dfrac{\pi}{2}< \pi\\0<\dfrac{\pi}{6}+ k2\pi < \pi\\0 <\dfrac{5\pi}{6} + k2\pi < \pi\end{array}\right.$
$\to \left[\begin{array}{l}-\dfrac{1}{2} < k< \dfrac{3}{2}\\-\dfrac{1}{2}<k< \dfrac{5}{12}\\-\dfrac{5}{12} <k< \dfrac{1}{6}\end{array}\right.$
$\to \left[\begin{array}{l}k = 0;\, 1\\k =0\\k = 0\end{array}\right.$
$\to \left[\begin{array}{l}x = \dfrac{\pi}{4}\\x = \dfrac{3\pi}{4}\\x = \dfrac{\pi}{6}\\x =\dfrac{5\pi}{6}\end{array}\right.$
