If [2 ^3 x+ 2 x x+4 x-4=0] has exactly 3 solutions in the interval [[0, n 2], n N], then t…

Question

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

Accepted Answer

$$\begin{aligned}
& 2 \sin ^3 x+2 \sin x \cdot \cos ^2 x+4 \sin x-4=0 \
& 2 \sin ^3 x+2 \sin x \cdot\left(1-\sin ^2 x\right)+4 \sin x-4=0 \
& 6 \sin x-4=0 \
& \sin x=\frac{2}{3} \
& \mathbf{n}=5 \text { (in the given interval) } \
& x^2+5 x+2=0 \
& x=\frac{-5 \pm \sqrt{17}}{2} \
& \text { Required interval }(-\infty, 0)
\end{aligned}$$

Solutions · 1