$\frac{\pi}{2}$ is correct, yes.
Proof: We want to use the following theorem: $\mathbb{E}(X|\mathcal{F}) = Y \Leftrightarrow \forall G \in \mathcal{G}: \int_G Y \, d\mathbb{P} = \int_G X \, d\mathbb{P} \qquad (\ast)$
where $G$ is a generator of the $\sigma$-Algebra $\mathcal{F}$. In this case we can choose $\mathcal{G} = \{\{\sin X \leq y\}; y \in \mathbb{R}\}$. Let $G = \{\sin X \leq y\} \in \mathcal{G}$, $y \in [0,1]$, then
$\int_G X \, d\mathbb{P} = \frac{1}{\pi} \int_0^\pi x \cdot 1_{(-\infty,y]}(\sin x) \, dx = \frac{1}{\pi} \cdot \left( \int_0^{\arcsin y} x \,dx + \int_{\pi-\arcsin y}^{\pi} x \, dx \right) \\ = \ldots = \arcsin y = \ldots = \frac{1}{\pi} \cdot \int_0^\pi \underbrace{\frac{\pi}{2}}_{=:Y} 1_{(-\infty,y]}(x) \, dx = \int_G \frac{\pi}{2}$
Clearly the equality holds also for $y \in \mathbb{R} \backslash [0,1]$. From $(\ast)$ follows $\mathbb{E}(X|\sin X)=\frac{\pi}{2}$.