3. In $\triangle A B C$, if $\sin C-\cos C+\sin B=\sqrt{3}$, and $A \geqslant \frac{\pi}{4}$, then $\sin B=$
3. $\frac{\sqrt{6}}{3}$ Detailed explanation: $B+C=\pi-A \leqslant \frac{3 \pi}{4}, C-\frac{\pi}{4} \leqslant \frac{\pi}{2}-B$, so $\sin \left(C-\frac{\pi}{4}\right) \leqslant \sin \left(\frac{\pi}{2}-B\right)=\cos B$, $\sqrt{3}=\sqrt{2} \sin \left(C-\frac{\pi}{4}\right)+\sin B \leqslant \sqrt{2} \cos B+\sin B=\sqrt{3} \sin (B+\varphi) \leqslant \sqrt{3}$, therefore $A=\frac{\pi}{4}$ and $\sin (B+\varphi)=1, \sin B=\frac{\sqrt{6}}{3}$.