Prove that

$3x-{x}^{3}<\frac{2}{\mathrm{sin}2x},\mathrm{\forall}x\in (0,\frac{\pi}{2})$

I have tried by proving that

$3x-{x}^{3}<\frac{9}{5\pi}x+\frac{3}{2}<\frac{2}{\mathrm{sin}2x},\mathrm{\forall}x\in (0,\frac{\pi}{2})$

with Jensen's inequality, but I hoped this problem would have a simpler solution.

$3x-{x}^{3}<\frac{2}{\mathrm{sin}2x},\mathrm{\forall}x\in (0,\frac{\pi}{2})$

I have tried by proving that

$3x-{x}^{3}<\frac{9}{5\pi}x+\frac{3}{2}<\frac{2}{\mathrm{sin}2x},\mathrm{\forall}x\in (0,\frac{\pi}{2})$

with Jensen's inequality, but I hoped this problem would have a simpler solution.