10. Solve the system of equations $\left\{\begin{array}{l}x+y=\frac{\pi}{2} \\ \cos 2 x=\cos x+\cos y\end{array}\right.$
10. From (1) we know $\cos y=\sin x$, substituting into (2) we get $\cos 2 x=\cos x+\sin x, (\cos x+\sin x)(\cos x-\sin x-1)=0$. From $\cos x+\sin x=0$ we get $\tan x=-1$, thus we have
$\left\{\begin{array}{l}x=k \pi-\frac{\pi}{4} \\ y=\frac{3 \pi}{4}-k \pi\end{array}\right.$ ( $k$ is an integer), then $\cos x-\sin x-1=0$, we get $\sin \left(\frac{\pi}{4}-x\right)=\frac{\sqrt{2}}{2}$, thus we have $\left\{\begin{array}{l}x=-2 k \pi \\ y=2 k \pi+\frac{\pi}{2},\end{array}\left\{\begin{array}{l}x=-2 k \pi-\frac{\pi}{2} \\ y=(2 k+1) \pi .\end{array}\right.\right.$ Therefore, the solutions to the original equation are $\left\{\begin{array}{l}x=k \pi-\frac{\pi}{4} \\ y=-k \pi+\frac{3 \pi}{4},\end{array} \quad\left\{\begin{array}{l}x=-2 k \pi \\ y=2 k \pi+\frac{\pi}{2},\end{array}\right.\right.$ $\left\{\begin{array}{l}x=-2 k \pi-\frac{\pi}{2} \\ y=(2 k+1) \pi,\end{array}\right.$ where $k$ is an integer.