`sin(x/2-π/3)=-\sqrt{3}/2`
`⇔sin(x/2-π/3)=sin(-π/3)`
`⇔` $\left[\begin{matrix} \dfrac{x}{2}-\dfrac{π}{3}=-\dfrac{π}{3}+k2π\\\dfrac{x}{2}-\dfrac{π}{3}=π+\dfrac{π}{3}+k2π\end{matrix}\right.$
`⇔` $\left[\begin{matrix} \dfrac{x}{2}=\dfrac{π}{3}-\dfrac{π}{3}+k2π\\\dfrac{x}{2}=\dfrac{π}{3}+π+\dfrac{π}{3}+k2π\end{matrix}\right.$
`⇔` $\left[\begin{matrix}\dfrac{x}{2}=k2π\\\dfrac{x}{2}=\dfrac{5π}{3}+k2π\end{matrix}\right.$
`⇔` $\left[\begin{matrix}x=k4π\\x=\dfrac{10π}{3}+k4π\end{matrix}\right.$
$\text{Vậy phương trình có 2 họ nghiệm}$
$\left[\begin{matrix}x=k4π\\x=\dfrac{10π}{3}+k4π\end{matrix}\right.$
