If 2sin^3x + sin2xcosx + 4sinx−4 = 0 has exactly 3 solutions in the interval [0, nπ/2], n ∈ N,

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 \(x^{2}+n x+(n-3)=0\) belong to :

(1) \((0, \infty)\)

(2) \((-\infty, 0)\)

(3) \(\left(-\frac{\sqrt{17}}{2}, \frac{\sqrt{17}}{2}\right)\)

(4) Z

Answer 1

Correct option is (2) \((-\infty, 0)\)

\(2 \sin ^{3} \mathrm{x}+2 \sin \mathrm{x} \cdot \cos ^{2} \mathrm{x}+4 \sin \mathrm{x}-4=0\)

\(2 \sin ^{3} \mathrm x+2 \sin\mathrm x \cdot\left(1-\sin ^{2}\mathrm x\right)+4 \sin \mathrm x-4=0\)

\(6 \sin \mathrm{x}-4=0\)

\(\sin \mathrm{x}=\frac{2}{3}\)

\(\mathrm n = 5\) (in the given interval)

\(\mathrm{x}^{2}+5 \mathrm{x}+2=0\)

\(\mathrm{x}=\frac{-5 \pm \sqrt{17}}{2}\)

Required interval \((-\infty, 0)\)