6. 62 Given that if $y=\sin x,-\frac{\pi}{2}<x<\frac{\pi}{2}$, then $x=\sin ^{-1} y$, now if $y=\sin x,\left(1992+\frac{1}{2}\right) \pi \leqslant x \leqslant\left(1993+\frac{1}{2}\right) \pi$, try to express $x$ in terms of $y$.
[Solution] $-\frac{\pi}{2} \leqslant 1993 \pi-x \leqslant \frac{\pi}{2}$,
and $\sin (1993 \pi-x)=\sin x=y$,
thus $x=1993 \pi-\sin ^{-1} y$.