For how many values of $x$ in $[0,\pi]$ is $\sin^{-1}(\sin8x)=\cos^{-1}(\cos x)$?

1. Since $\cos^{-1}(\cos x)=x$ on $[0,\pi]$, and $\sin^{-1}(\sin\theta)\in[-\pi/2,\pi/2]$, we need $x\in[0,\pi/2]$ and $\arcsin(\sin8x)=x$.
2. With $\theta=8x\in[0,4\pi]$: $\arcsin(\sin\theta)$ equals $\theta$ on $[0,\frac\pi2]$, $\pi-\theta$ on $[\frac\pi2,\frac{3\pi}2]$, $\theta-2\pi$ on $[\frac{3\pi}2,\frac{5\pi}2]$, $3\pi-\theta$ on $[\frac{5\pi}2,\frac{7\pi}2]$, and $\theta-4\pi$ on $[\frac{7\pi}2,4\pi]$.
3. The equations $8x=x$, $\pi-8x=x$, $8x-2\pi=x$, $3\pi-8x=x$ give $x=0,\frac\pi9,\frac{2\pi}7,\frac\pi3$, and each $\theta=8x$ lies in the corresponding piece. The last piece would require $x=\frac{4\pi}7>\frac\pi2$, invalid.
4. There are $\boxed4$ solutions.