sin 3x = cos 2x \Rightarrow $$$$\,\,\, 3 sin x $-$ 4 sin 3 x = 1 $-$ 2 sin 2 x \Rightarrow $$$$\,\,\, 4 sin 3 x $-$ 2 sin 2 x $-$ 3 sin x + 1 = 0 \Rightarrow $$$$\,\,\, sin x = 1, ${{ - 2 \pm 2\sqrt 5 } \over 8}$ In the interval $\left( {{\pi \over 2},\pi } \right)$, sin x = ${{ - 2 + 2\sqrt 5 } \over 8}$ So, there is only one solution.