The maximum value of $f\left(x\right)=2\sin x+\sin 2x\text{, }$in the interval $\left[0,\frac{3\pi }{2}\right]$ is

$\begin{array}{l}f\left(x\right)=2\sin x+\sin 2x=2\sin x+2\sin x\cos x\\ =2\sin x(1+\cos x)\\ f^{\mathrm{\prime }}\left(x\right)=2\cos x(1+\cos x)-2\sin x\sin x\\ =2\left[\cos x+{\cos }^{2}x-\left(1-{\cos }^{2}x\right)\right]\\ =2\left[2{\cos }^{2}x+\cos x-1\right]\\ =2(\cos x+1)(2\cos x-1)\end{array}$

Now, $f^{\mathrm{\prime }}\left(x\right)=0\Rightarrow x=\pi /3,\pi (\because 0\le x\le 3\pi /2)$

we have $f\left(0\right)=0,$ of $\left(\frac{3\pi }{2}\right)=-2$

$f\left(\frac{\pi }{3}\right)=2\left(\frac{\sqrt{3}}{2}\right)+\frac{\sqrt{3}}{2}=\frac{3}{2}\sqrt{3}$

and $f\left(\pi \right)=0,f\left(\frac{3\pi }{2}\right)=-2$

Thus, maximum value is $\frac{3}{2}\sqrt{3}$