Given: $2{\sin }^{3}x+\sin 2x\cos x+4\sin x-4=0$

$\Rightarrow 2{\sin }^{3}x+2\sin x·{\cos }^{2}x+4\sin x-4=0$

$\Rightarrow 2{\sin }^{3}x+2\sin x·\left(1-{\sin }^{2}x\right)+4\sin x-4=0$

$\Rightarrow 2{\sin }^{3}x+2\sin x-2{\sin }^{3}x+4\sin x-4=0$

$\Rightarrow 6\sin x-4=0$

$\Rightarrow \sin x=\frac{2}{3}$

Now, for exactly three solution we get,

$\Rightarrow n=5$ (in the given interval)

So, $x^2+nx+\left(n-3\right)=0$

$\Rightarrow x^2+5x+2=0$

$\Rightarrow x=\frac{-5\pm \sqrt{17}}{2}$

So, the required interval is $(-\infty ,0)$.