3. Let $\alpha, \beta \in(0, \pi), \cos \alpha, \cos \beta$ be the roots of the equation $5 x^{2}-3 x-1=0$, then the value of $\sin \alpha \sin \beta$ is $\qquad$ .
Answer: $\frac{\sqrt{7}}{5}$.
Solution: Given $\cos \alpha+\cos \beta=\frac{3}{5}, \cos \alpha \cos \beta=-\frac{1}{5}$, thus
$$
\begin{array}{c}
(\sin \alpha \sin \beta)^{2}=\left(1-\cos ^{2} \alpha\right)\left(1-\cos ^{2} \beta\right)=1-\cos ^{2} \alpha-\cos ^{2} \beta+\cos ^{2} \alpha \cos ^{2} \beta \\
=(1+\cos \alpha \cos \beta)^{2}-(\cos \alpha+\cos \beta)^{2}=\left(\frac{4}{5}\right)^{2}-\left(\frac{3}{5}\right)^{2}=\frac{7}{25} .
\end{array}
$$

Also, since $\alpha, \beta \in(0, \pi)$, we know $\sin \alpha \sin \beta>0$, thus $\sin \alpha \sin \beta=\frac{\sqrt{7}}{5}$.