9. Let $\alpha, \beta, \gamma$ satisfy $0<\alpha<\beta<\gamma<2 \pi$. If for any $x \in \mathbf{R}, \cos (x+\alpha)+\cos (x+$ $\beta)+\cos (x+\gamma)=0$, then $\gamma-\alpha=$ $\qquad$ .
Let $x=0, \frac{\pi}{2}$, we get $\left\{\begin{array}{l}\cos \alpha+\cos \beta+\cos \gamma=0, \\ \sin \alpha+\sin \beta+\sin \gamma=0\end{array} \Rightarrow\left\{\begin{array}{l}\cos \alpha+\cos \gamma=-\cos \beta, \\ \sin \alpha+\sin \gamma=-\sin \beta\end{array}\right.\right.$ $\Rightarrow 2+2 \cos (\gamma-\alpha)=1 \Rightarrow \gamma-\alpha=\frac{2 \pi}{3}$ or $\frac{4 \pi}{3}$. Similarly, $\beta-\alpha=\frac{2 \pi}{3}$ or $\frac{4 \pi}{3}$, $\gamma-\beta=\frac{2 \pi}{3}$ or $\frac{4 \pi}{3}$. Combining the conditions, we get $\gamma-\beta=\beta-\alpha=\frac{2 \pi}{3}, \gamma-\alpha=\frac{4 \pi}{3}$. At this point, $\cos (x+\alpha)+\cos (x+\beta)+\cos (x+\gamma)$ $=\cos (x+\alpha)+\cos \left(x+\alpha+\frac{2 \pi}{3}\right)+\cos \left(x+\alpha+\frac{4 \pi}{3}\right)$ $\xlongequal{\theta=x+\alpha} \cos \theta+\cos \left(\theta+\frac{2 \pi}{3}\right)+\cos \left(\theta+\frac{4 \pi}{3}\right)$ $=2 \cos \left(\theta+\frac{2 \pi}{3}\right) \cos \frac{2 \pi}{3}+\cos \left(\theta+\frac{2 \pi}{3}\right)=0$. Therefore, $\gamma-\alpha=\frac{4 \pi}{3}$.