1. Simplify the trigonometric equation

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

Use $\sin 2x=2\sin x\cos x$ so $\sin 2x\cos x=2\sin x\cos^2 x$

Hence the equation becomes $2\sin^3 x+2\sin x\cos^2 x+4\sin x-4=0$

Factor $2\sin x$ from the first two terms: $2\sin x(\sin^2 x+\cos^2 x)+4\sin x-4=0$ Since $\sin^2 x+\cos^2 x=1$, $2\sin x+4\sin x-4=0$ $6\sin x-4=0$ $\sin x=\frac{2}{3}$

2. Count the number of solutions in $\left[0,\frac{n\pi}{2}\right]$

Let $\alpha=\sin^{-1}\left(\frac23\right)$ where $0<\alpha<\frac{\pi}{2}$.

General solutions of $\sin x=\frac23$ are $x=2k\pi+\alpha \quad \text{or} \quad x=2k\pi+(\pi-\alpha), \qquad k\in \mathbb Z$

Now count solutions in intervals of the form $\left[0,\frac{n\pi}{2}\right]$.

• In $[0,\frac{\pi}{2}]$: only $x=\alpha$ appears, so count $=1$.
• In $[0,\pi]$: both $\alpha$ and $\pi-\alpha$ appear, so count $=2$.
• In $[0,\frac{3\pi}{2}]$: still only those two, so count $=2$.
• In $[0,2\pi]$: still count $=2$.
• In $[0,\frac{5\pi}{2}]$: additionally $2\pi+\alpha$ appears, so count $=3$.

Thus the smallest natural number $n$ for which the interval contains exactly 3 solutions is $\frac{n\pi}{2}=\frac{5\pi}{2} \implies n=5$

3. Solve the quadratic equation

Now consider $x^2+nx+(n-3)=0$ With $n=5$, $x^2+5x+2=0$

Using the quadratic formula, $x=\frac{-5\pm\sqrt{25-8}}{2}=\frac{-5\pm\sqrt{17}}{2}$

So the roots are $\frac{-5+\sqrt{17}}{2}, \qquad \frac{-5-\sqrt{17}}{2}$

Since $\sqrt{17}<5$, both roots are negative. Therefore both roots belong to $(-\infty,0)$

4. Check options

• A: $(0,\infty)$ — false
• B: $\mathbb Z$ — false
• C: $\left(-\frac{\sqrt{17}}{2},\frac{\sqrt{17}}{2}\right)$ — false, because one root is less than $-\frac{\sqrt{17}}{2}$
• D: $(-\infty,0)$ — true

Hence the correct option is D.