How many solutions does $\sin\left(\frac{2\pi}3\cos x\right)=\cos\left(\frac{2\pi}3\sin x\right)$ have in $[0,\pi]$?

1. Write the right side as $\sin\left(\frac\pi2-\frac{2\pi}3\sin x\right)$. Then either $\frac{2\pi}3\cos x=\frac\pi2-\frac{2\pi}3\sin x+2k\pi$ or $\frac{2\pi}3\cos x=\frac\pi2+\frac{2\pi}3\sin x+2k\pi$, i.e. $\cos x+\sin x=\frac34+3k$ or $\cos x-\sin x=\frac34+3k$.
2. On $[0,\pi]$, $\cos x+\sin x=\sqrt2\sin\left(x+\frac\pi4\right)\in[-1,\sqrt2]$ and $\cos x-\sin x=\sqrt2\cos\left(x+\frac\pi4\right)\in[-\sqrt2,1]$, so only $k=0$ is possible.
3. $\sin\left(x+\frac\pi4\right)=\frac3{4\sqrt2}\approx0.53$ with $x+\frac\pi4\in\left[\frac\pi4,\frac{5\pi}4\right]$ has exactly one solution ($x\approx1.80$), since $0.53<\sin\frac\pi4$. And $\cos\left(x+\frac\pi4\right)=\frac3{4\sqrt2}$ has exactly one solution ($x\approx0.23$).
4. The answer is $\boxed2$.