11. Let $0<\alpha<\pi, \pi<\beta<2 \pi$. If for any $x \in \mathbf{R}$, we have
$$
\cos (x+\alpha)+\sin (x+\beta)+\sqrt{2} \cos x=0
$$

always holds, find the value of $\alpha \beta$.
Given the equation can be transformed into
$$
(\cos \alpha+\sin \beta+\sqrt{2}) \cos x+(\cos \beta \cdots \sin \alpha) \sin x=0 \text {. }
$$

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

Squaring and adding, then simplifying, we get $\cos \alpha=-\frac{\sqrt{2}}{2}$.
Since $0<\alpha<\pi$, we have $\alpha=\frac{3 \pi}{4}$.
Thus, $\cos \beta=\sin \alpha=\frac{\sqrt{2}}{2}$.
Since $\pi<\beta<2 \pi$, we have $\beta=\frac{7 \pi}{4}$.