Answer
$c=\pi$
Work Step by Step
Here, $f(x)=\sin(x/2)$ in interval $[\pi/2,3\pi/2]$.
1. $f(x)$ is continuous over the given interval $[\pi/2,3\pi/2]$.
2. Similarly, since $f(x)$ is a function of sine, it is differentiable on real values and the open interval $(\pi/2,3\pi/2)$.
3. $f(\pi/2)=\sin(\pi/4)=\frac{1}{\sqrt2}$. Similarly,$f(3\pi/2)=\sin(3\pi/4)=\sin(\pi/2+\pi/4)=\sin(\pi/4)=\frac{1}{\sqrt2}$. Thus, $f(\pi/2)=f(3\pi/2)$.
Thus, all three criteria of Rolle's theorem satisfied. This implies that there exists a number $c$ in the open interval $(\pi/2,3\pi/2)$ such that $f'(c)=0$. Let's differentiate $f(x)$, $$f'(x)=\frac12\cos\bigg(\frac x2\bigg)\implies\frac12\cos\bigg(\frac c2\bigg)=0$$ $$\frac c2=\frac \pi2, 3\pi/2\implies c=\pi,3\pi.$$ Thus, $c=\pi$ satisfies the required condition.