4. If $\alpha \_\beta$ are acute angles, and
$$
\begin{array}{l}
3 \sin ^{2} \alpha+2 \sin ^{2} \beta=1, \\
3 \sin 2 \alpha-2 \sin 2 \beta=0,
\end{array}
$$

then $\sin (\alpha+2 \beta)=(\quad)$.
(A) $\frac{1}{2}$
(B) $\frac{\sqrt{2}}{2}$
(C) $\frac{\sqrt{3}}{2}$
(D) 1
4.D.
$$
\begin{array}{l}
\text { Given } \cos (\alpha+2 \beta)=\cos \alpha \cdot \cos 2 \beta-\sin \alpha \cdot \sin 2 \beta \\
=\cos \alpha \cdot 3 \sin ^{2} \alpha-\sin \alpha \cdot 3 \sin \alpha \cdot \cos \alpha=0,
\end{array}
$$

and $0<\alpha+2 \beta<\frac{3 \pi}{2}$, we have $\alpha+2 \beta=\frac{\pi}{2}$. Therefore, $\sin (\alpha+2 \beta)=1$.