Đáp án:
$x = \pi + k4\pi\,\,\,\,\,(k \in \Bbb Z)$
Giải thích các bước giải:
$\begin{array}{l}\sin^3\dfrac{x}{2}-3\sin\dfrac{x}{2}+2 =0\\ \Leftrightarrow \sin^3\dfrac{x}{2} - \sin^2\dfrac{x}{2} + \sin^2\dfrac{x}{2} - \sin\dfrac{x}{2} -2\sin\dfrac{x}{2}+2=0\\ \Leftrightarrow \sin^2\dfrac{x}{2}\left(\sin\dfrac{x}{2}-1\right)+\sin\dfrac{x}{2}\left(\sin\dfrac{x}{2}-1\right) -2\left(\sin\dfrac{x}{2}-1\right)=0\\ \Leftrightarrow \left(\sin\dfrac{x}{2}-1\right)\left(\sin^2\dfrac{x}{2}+ \sin\dfrac{x}{2} -2\right)=0\\ \Leftrightarrow \left(\sin\dfrac{x}{2}-1\right)^2\cdot\left(\sin\dfrac{x}{2}+2\right)=0\\ \Leftrightarrow \left[\begin{array}{l}\sin\dfrac{x}{2}-1 = 0\\\sin\dfrac{x}{2}+2=0\end{array}\right.\\ \Leftrightarrow \left[\begin{array}{l}\sin\dfrac{x}{2}=1\\\sin\dfrac{x}{2}=-2 < -1 &(loại)\end{array}\right.\\ \Leftrightarrow \dfrac{x}{2} = \dfrac{\pi}{2} + k2\pi\\ \Leftrightarrow x = \pi + k4\pi\,\,\,\,\,(k \in \Bbb Z)\end{array}$
