smc

Define a new number $B$ such that $\cos (B)=\frac{1}{2}$ and $\sin (B)=-\frac{\sqrt{3}}{2}$. Notice that $B=\frac{5\pi }{3}$. Therefore, we have $\sin \frac{A}{2}-\sqrt{3}\cos \frac{A}{2}=2\cos (B)\sin \left(\frac{A}{2}\right)+2\sin (B)\cos \left(\frac{A}{2}\right)$. Recognizing the angle sum formula, we can rewrite this as $\sin \frac{A}{2}-\sqrt{3}\cos \frac{A}{2}=2\sin \left(\frac{A}{2}+B\right)$. We can now clearly see that the minimum is $-2$, and this is achieved when $\frac{A}{2}+B=\frac{3\pi }{2}$. Plugging in $B=\frac{5\pi }{3}$, we get $\frac{A}{2}=-\frac{\pi }{6}⟹A=-\frac{\pi }{3}=-60^{\circ }$. $\boxed{\text{none of these}}$