8. $\alpha, \beta \in(0, \pi)$, and $\cos \alpha+\cos \beta-\cos (\alpha+\beta)=\frac{3}{2}$, then $2 \alpha+\beta=$
8. $\pi$.

The equation can be transformed into $4 \cos ^{2} \frac{\alpha+\beta}{2}-4 \cos \frac{\alpha-\beta}{2} \cos \frac{\alpha+\beta}{2}+1=0$, and $\cos \frac{\alpha+\beta}{2}$ is a real root of the quadratic equation $4 x^{2}-4 \cos \frac{\alpha-\beta}{2} x+1=0$. Therefore, $\Delta=\left(-4 \cos \frac{\alpha-\beta}{2}\right)^{2}-4 \times 4 \times 1 \geqslant 0 \Leftrightarrow\left|\cos \frac{\alpha-\beta}{2}\right| \geqslant 1$. But $\left|\cos \frac{\alpha-\beta}{2}\right| \leqslant 1$, so $\cos \frac{\alpha-\beta}{2}= \pm 1$. Since $\alpha, \beta \in(0, \pi) \Leftrightarrow-\frac{\pi}{2}<\frac{\alpha-\beta}{2}<\frac{\pi}{2}$, it follows that $\frac{\alpha-\beta}{2}=$ $0 \Leftrightarrow \alpha=\beta$. Substituting into the original equation, we get $2 \cos \alpha-\cos 2 \alpha=1 \Rightarrow \cos \alpha=\frac{1}{2} \Rightarrow \alpha=\frac{\pi}{3}$. Therefore, $2 \alpha+\beta=\pi$.