Taking cosine gives \[\sin^2x+\frac32\sin x=\cos\frac{2\pi}{3}=-\frac12.\] Thus \[2\sin^2x+3\sin x+1=(2\sin x+1)(\sin x+1)=0.\] The possibilities are \(\sin x=-1/2\) and \(\sin x=-1\). In the stated interval, \[\boxed{x=\frac{7\pi}{6},\frac{3\pi}{2},\frac{11\pi}{6}}.\] For each root, the inverse-cosine argument is \(-1/2\), whose principal value is \(2\pi/3\).