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

Given data: $f\left(x\right)=2\sin x+\sin 2x$

Differentiating with respect to $x, f’\left(x\right)=2\cos x+2\cos 2x$

$\Rightarrow f’\left(x\right)=2(cosx+2co{s}^{2}x-1)[\because cos2x=2co{s}^{2}x-1]$

$\Rightarrow f\text{'}\left(x\right)=2(2cosx-1)(cosx+1)$

To find critical point of $f\left(x\right), f\text{'}\left(x\right)=0$

$cosx=-1,\frac{1}{2}\Rightarrow x=\pi ,\frac{\pi }{3}$

The values at critical points \lambda end points at $x=0,\frac{\pi }{3},\pi ,\frac{3\pi }{2}$

$$\begin{gathered} \Rightarrow f\left(0\right)=0 \\ f\left(\frac{\pi }{3}\right)=2sin\frac{\pi }{3}+sin\left(2\frac{\pi }{3}\right)=\frac{\sqrt{3}}{2}+\frac{2\sqrt{3}}{2} \\ \Rightarrow f\left(\frac{\pi }{3}\right)=\frac{3\sqrt{3}}{2} \\ \Rightarrow f\left(\pi \right)=0 \\ f\left(\frac{3\pi }{2}\right)=2sin\frac{3\pi }{2}+sin\left(2\frac{3\pi }{2}\right) \\ \Rightarrow f\left(\frac{3\pi }{2}\right)=-2 \end{gathered}$$

Hence the greatest value of the function $f\left(x\right)=2\sin x+\sin 2x$ in the critical interval $\left[0,\frac{{\displaystyle 3\pi }}{{\displaystyle 2}}\right]$ is $\frac{3\sqrt{3}}{2}$.