Express $\sin (2x)$ in the form $\dfrac{a\pi^2+b\pi + c}{d},$ where $a,b,c,d$ are integers 
Let $0<x<\dfrac{\pi}{2}.$ If $x$ is such that $\cos\left(\dfrac{3}{2}\cos x\right) = \sin\left(\dfrac{3}{2}\sin x\right),$ then express $\sin\, (2x)$ in the form $\dfrac{a\pi^2+b\pi + c}{d},$ where $a,b,c,d$ are integers.

Here are my proofs that $\sin \left(\dfrac{\pi}{2}-x\right)=\cos\left(x\right).$ First, angle addition says that $\sin\left( \dfrac{\pi}{2}-x\right)=\sin\dfrac{\pi}{2}\cos x-\sin x\cos\dfrac{\pi}{2}=(1)\cos x - \sin x (0) = \cos x.$ Second, a geometric proof. Consider a right triangle. We may assume WLOG that the hypotenuse is $1$. If not, then multiply all sides by the multiplicative inverse of the hypotenuse so that it is (the triangle obtained is similar, so the angles are preserved). Let one of the acute angles be $\alpha$. Then the side adjacent to that angle has length $\cos \alpha$. Now consider the angle $\dfrac{\pi}{2} - \alpha$. The side opposite to this angle has length $\cos \alpha$. But since $\sin x = \dfrac{\text{opposite}}{\text{hypotenuse}}, \sin\left(\dfrac{\pi}{2} - \alpha\right)=\cos \alpha.$ Now, if $\alpha$ is not acute, we may add an integer multiple of $2\pi$ to $\alpha$ so that it is, without changing the value of $\sin \alpha$ since $\sin x$ is $2\pi$ periodic.
Now, since $0<x < \dfrac{\pi}{2}, \sin x,\cos x \in (0, 1)\Rightarrow \dfrac{3}{2}\sin x, \dfrac{3}{2}\cos x \in (0,\dfrac{3}{2})\subseteq (0,\dfrac{\pi}{2})$. So, using this fact, we have that $\cos \left(\dfrac{3}{2}\cos x\right)=\sin \left(\dfrac{3}{2}\sin x\right)\Leftrightarrow\dfrac{3}{2}\cos x+\dfrac{3}{2}\sin x = \dfrac{\pi}{2}.$ Using the fact that $\sin x = \sqrt{1-\cos ^2 x},$ I get that $\sin x = \dfrac{\frac{2\pi}{3}\pm \frac{2}{3}\sqrt{18-\pi^2}}{4},$ but I'm not sure how to use this to get $\sin \,(2x)$ into the desired form.

You can use the fact that
$$\begin{align}\dfrac{3}{2}\cos x+\dfrac{3}{2}\sin x = \dfrac{\pi}{2}&\Leftrightarrow \sin x + \cos x = \dfrac{\pi}{3}\\
&\Leftrightarrow \cos x = \dfrac{\pi}{3}-\sin x.\end{align}$$
Since $\sin (2x) = 2\sin x \cos x$, we have $\sin (2x) = 2\cdot\dfrac{\frac{2\pi}{3}+ \frac{2}{3}\sqrt{18-\pi^2}}{4}\cdot \left(\dfrac{\pi}{3} - \dfrac{\frac{2\pi}{3}+ \frac{2}{3}\sqrt{18-\pi^2}}{4}\right)= \dfrac{\pi^2-9}{9}.$
