5 Given $\alpha, \beta$ are acute angles, and $3 \sin ^{2} \alpha+2 \sin ^{2} \beta=1, 3 \sin 2 \alpha-2 \sin 2 \beta=0$, then $\alpha+2 \beta=$

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$5 \frac{\pi}{2}$.
$$
\frac{3(1-\cos 2 \alpha)}{2}+1-\cos 2 \beta=1
$$

So
$$
3 \cos 2 \alpha+2 \cos 2 \beta=3
$$

Rearranging and squaring gives
$$
9 \cos ^{2} 2 \alpha=9+4 \cos ^{2} 2 \beta-12 \cos 2 \beta
$$

Also
$$
9 \sin ^{2} 2 \alpha=4 \sin ^{2} 2 \beta
$$

Adding and simplifying gives
$$
\cos 2 \beta=\frac{1}{3}
$$

So $\sin 2 \beta=\frac{2 \sqrt{2}}{3}, \cos 2 \alpha=\frac{7}{9}$.
$$
\begin{array}{c}
\sin \alpha=\sqrt{\frac{1-\cos 2 \alpha}{2}}=\frac{1}{3}=\cos 2 \beta . \\
\alpha+2 \beta=\frac{\pi}{2} .
\end{array}
$$