If $$2\sin^3 x + \sin 2x \cos x + 4\sin x - 4 = 0$$ has exactly $$3$$ solutions in the interval $$\left[0, \frac{n\pi}{2}\right]$$, $$n \in \mathbb{N}$$, then the roots of the equation $$x^2 + nx + (n - 3) = 0$$ belong to :

Let $$\alpha = \arcsin\!\left(\dfrac{2}{3}\right)$$.
Given $$2\sin^{3}x+\sin 2x \cos x+4\sin x-4=0$$.

Use $$\sin 2x = 2\sin x \cos x$$ and $$\sin^{2}x = 1-\cos^{2}x$$:

$$2\sin^{3}x = 2\sin x(1-\cos^{2}x)=2\sin x-2\sin x\cos^{2}x$$

$$\sin 2x \cos x = 2\sin x\cos^{2}x$$

Adding these inside the equation, the $$\pm2\sin x\cos^{2}x$$ terms cancel:

$$\bigl(2\sin x-2\sin x\cos^{2}x\bigr)+2\sin x\cos^{2}x+4\sin x-4=0$$

$$6\sin x-4=0 \;\;\Longrightarrow\;\; \sin x=\dfrac{2}{3}$$

General solutions of $$\sin x=\dfrac{2}{3}$$ are

$$x = \alpha + 2k\pi \quad\text{and}\quad x = \pi-\alpha+2k\pi,\;k\in\mathbb{Z}$$

$$x_1 = \alpha$$

$$x_2 = \pi-\alpha$$

$$x_3 = 2\pi+\alpha$$

$$x_4 = 3\pi-\alpha$$

(and so on, two roots in every length-$$2\pi$$ interval).

The interval upper end is $$L=n\dfrac{\pi}{2}$$. To have exactly three solutions we need

$$2\pi+\alpha \;\le\;L\;\lt\;3\pi-\alpha\;-(1)$$

$$2\pi+\alpha \approx 6.2832+0.7297 = 7.0129$$

$$3\pi-\alpha \approx 9.4248-0.7297 = 8.6951$$

The factor $$\dfrac{\pi}{2}\approx1.5708$$. Divide the inequalities in $$(1)$$ by $$\dfrac{\pi}{2}$$:

$$\dfrac{7.0129}{1.5708}\;\le\;n\;\lt\;\dfrac{8.6951}{1.5708}$$

$$4.463\;\le\;n\;\lt\;5.533$$

With $$n\in\mathbb{N}$$, the only possible value is

$$n = 5$$
Put $$n=5$$ in $$x^{2}+nx+(n-3)=0$$:

$$x^{2}+5x+2=0\;-(2)$$

Discriminant $$\Delta = 5^{2}-4\cdot1\cdot2 = 25-8 = 17$$

Roots of $$(2)$$ are

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

$$x_{1} = \dfrac{-5+\sqrt{17}}{2}\approx\dfrac{-5+4.123}{2}\approx-0.438$$

$$x_{2} = \dfrac{-5-\sqrt{17}}{2}\approx\dfrac{-5-4.123}{2}\approx-4.562$$

Both roots are negative, so they lie in the interval $$(-\infty,0)$$.