[Algebra]
Problem: Suppose that $x$ satisfies $0<x<\frac{\pi}{2}$ and $\cos \left(\frac{3}{2} \cos x\right)=\sin \left(\frac{3}{2} \sin x\right)$.

Determine all possible values of $\sin 2 x$, expressing your answers in the form $\frac{a \pi^{2}+b \pi+c}{d}$ where $a, b, c, d$ are integers.

Solution: Since $0<x<\frac{\pi}{2}$, then $0<\cos x<1$ and $0<\sin x<1$.

This means that $0<\frac{3}{2} \cos x<\frac{3}{2}$ and $0<\frac{3}{2} \sin x<\frac{3}{2}$. Since $3<\pi$, then $0<\frac{3}{2} \cos x<\frac{\pi}{2}$ and $0<\frac{3}{2} \sin x<\frac{\pi}{2}$.

If $Y$ and $Z$ are angles with $0<Y<\frac{\pi}{2}$ and $0<Z<\frac{\pi}{2}$, then $\cos Y=\sin Z$ exactly when $Y+Z=\frac{\pi}{2}$. To see this, we could picture points $R$ and $S$ on the unit circle corresponding to the angles $Y$ and $Z$; the $x$-coordinate of $R$ is equal to the $y$-coordinate of $S$ exactly when the angles $Y$ and $Z$ are complementary.

Therefore, the following equations are equivalent:

$$
\begin{aligned}
\cos \left(\frac{3}{2} \cos x\right) & =\sin \left(\frac{3}{2} \sin x\right) \\
\frac{3}{2} \cos x+\frac{3}{2} \sin x & =\frac{\pi}{2} \\
\cos x+\sin x & =\frac{\pi}{3} \\
(\sin x+\cos x)^{2} & =\frac{\pi^{2}}{9} \\
\sin ^{2} x+2 \sin x \cos x+\sin ^{2} x & =\frac{\pi^{2}}{9} \\
2 \sin x \cos x+\left(\sin ^{2} x+\cos ^{2} x\right) & =\frac{\pi^{2}}{9} \\
\sin 2 x+1 & =\frac{\pi^{2}}{9} \\
\sin 2 x & =\frac{\pi^{2}-9}{9}
\end{aligned}
$$

Therefore, the only possible value of $\sin 2 x$ is $\frac{\pi^{2}-9}{9}$.

Final answer: $\frac{\pi^{2}-9}{9}$