1. Solve the equation
$$
\sin \sin x=\sin (\cos x+1)
$$
1. Consider two cases.
(1) If $\sin x=\cos x+1+2 \pi l(l \in \mathbf{Z})$, then
$$
\begin{array}{l}
|2 \pi l|=|-1+\sin x-\cos x| \leqslant 1+\sqrt{2} \\
\Rightarrow l=0 .
\end{array}
$$

Notice that,
$$
\begin{array}{l}
\sin x-\cos x=1 \\
\Leftrightarrow \sin \left(x-\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2} \\
\Leftrightarrow x=\frac{\pi}{4}+(-1)^{n} \frac{\pi}{4}+\pi n(n \in \mathbf{Z}) .
\end{array}
$$
(2) If $\sin x=\pi-\cos x-1+2 \pi l(l \in \mathbf{Z})$, then
$$
\pi \leqslant|\pi+2 \pi l| \leqslant 1+|\sin x+\cos x| \leqslant 1+\sqrt{2} \text {. }
$$

This is impossible.
In conclusion, $x=\frac{\pi}{4}+(-1)^{n} \frac{\pi}{4}+\pi n(n \in \mathbf{Z})$.