The number of real solutions in the interval $\left[-\frac{3\pi }{2},\frac{7\pi }{2}\right]$ of the equation $4{\sin }^{2}x-8\sin x+3=0$ is

$\begin{array}{l}4{\sin }^{2}x-8\sin x+3=0\Rightarrow 4{\sin }^{2}x-6\sin x-2\sin x+3=0\\ \Rightarrow (2\sin x-3)(2\sin x-1)=0\Rightarrow \sin x=\frac{1}{2}\Rightarrow \sin x=\sin \frac{\pi }{6}\Rightarrow x=n\pi +(-1{)}^n\frac{\pi }{6}\\ x\in \left[\frac{-3\pi }{2},\frac{7\pi }{2}\right]\Rightarrow x=\frac{-7\pi }{6},\frac{\pi }{6},\frac{5\pi }{6},\frac{13\pi }{6},\frac{17\pi }{6}.\therefore \text{ The number of real solutions }=5\end{array}$