2.1.13 ** 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 $f(x)=\cos (x+\alpha)+\cos (x+\beta)+\cos (x+\gamma)$, for $x \in \mathbf{R}, f(x) \equiv 0$, we have $f(-\alpha)=0, f(-\beta)=0, f(-\gamma)=0$, that is,
$$
\begin{array}{l}
\cos (\beta-\alpha)+\cos (\gamma-\alpha)=-1, \cos (\alpha-\beta)+\cos (\gamma-\beta)=-1, \\
\cos (\alpha-\gamma)+\cos (\beta-\gamma)=-1,
\end{array}
$$

Therefore, $\cos (\beta-\alpha)=\cos (\gamma-\beta)=\cos (\gamma-\alpha)=-\frac{1}{2}$.
Since $0<\alpha<\beta<\gamma<2 \pi$, we have $\beta-\alpha, \gamma-\alpha, \gamma-\beta \in\left\{\frac{2 \pi}{3}, \frac{4 \pi}{3}\right\}$. Also, $\beta-\alpha<\gamma-\alpha, \gamma-\beta<\gamma-\alpha$, so only $\beta-\alpha=\gamma-\beta=\frac{2 \pi}{3}, \gamma-\alpha=\frac{4 \pi}{3}$.

On the other hand, let $\beta=\alpha+\frac{2 \pi}{3}, \gamma=\alpha+\frac{4 \pi}{3}$, for any real number $x$, let $x+\alpha=\theta$, since the three points $(\cos \theta, \sin \theta),\left(\cos \left(\theta+\frac{2 \pi}{3}\right), \sin \left(\theta+\frac{2 \pi}{3}\right)\right),\left(\cos \left(\theta+\frac{4 \pi}{3}\right), \sin \left(\theta+\frac{4 \pi}{3}\right)\right)$ form the three vertices of an equilateral triangle on the unit circle, it is easy to see that $\cos \theta+\cos \left(\theta+\frac{2 \pi}{3}\right)+\cos \left(\theta+\frac{4 \pi}{3}\right)=0$, that is,
$$
\cos (x+\alpha)+\cos (x+\beta)+\cos (x+\gamma)=0 .
$$