(11) Let $0<\alpha<\pi, \pi<\beta<2 \pi$, if for any $x \in \mathbf{R}$, the equation
$$
\cos (x+\alpha)+\sin (x+\beta)+\sqrt{2} \cos x=0
$$

always holds, try to find the values of $\alpha, \beta$.
11 The given equation can be transformed into
$$
(\cos \alpha+\sin \beta+\sqrt{2}) \cos x+(\cos \beta-\sin \alpha) \sin x=0 .
$$

The necessary and sufficient condition for the above equation to hold for any $x \in \mathbf{R}$ is
$$
\left\{\begin{array} { l } 
{ \operatorname { c o s } \alpha + \operatorname { s i n } \beta + \sqrt { 2 } = 0 , } \\
{ \operatorname { c o s } \beta - \operatorname { s i n } \alpha = 0 , }
\end{array} \text { i.e., } \left\{\begin{array}{l}
\sin \beta=-\cos \alpha-\sqrt{2}, \\
\cos \beta=\sin \alpha .
\end{array}\right.\right.
$$

Squaring and adding, we get $(-\cos \alpha-\sqrt{2})^{2}+\sin ^{2} \alpha=1$. Simplifying, we obtain $\cos \alpha=-\frac{\sqrt{2}}{2}$. Since $0<\alpha<\pi$, we have $\alpha=\frac{3 \pi}{4}$, and thus $\cos \beta=\sin \alpha=\frac{\sqrt{2}}{2}$.
Since $\pi<\beta<2 \pi$, we have $\beta=\frac{7 \pi}{4}$.