If 2 sin 3 x+sin 2 x cos x+4 sin x-4=0 has exactly 3 solutions in the interval [0, frac n π 2 ], n in N , then the roots of the equation x 2 +n x+(n-3)=0 belong to :

(-frac √(17) 2 , frac √(17) 2 )

Mathematics · Trigonometry Ratios and Identities

JEE Main 2024 — 30 January, Shift 1 — Question 19

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 :

Option A:
$(0, \infty)$
Option B:
$(-\infty, 0)$
Correct
Option C:
$\left(-\frac{\sqrt{17}}{2}, \frac{\sqrt{17}}{2}\right)$
Option D:
Z

Answer: B

Step-by-step solution

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

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

$6 \sin \mathrm{x}-4=0$

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

$\mathbf{n}=\mathbf{5}$ (in the given interval) $x^{2}+5 x+2=0$

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

Required interval $(-\infty, 0)$

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 30 January, Shift 1
Subject: Mathematics
Chapter: Trigonometry Ratios and Identities
Topic: Periodicity of trigometric functions,Solutions of trigonometric equations