Problem 573

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