JEE Main 2024MathematicsTrigonometric EquationsHardMCQ

JEE Main 2024 — Trigonometric Equations Question with Solution

JEE Main 2024 (30 Jan Shift 1)

Question

If 2sin3x+sin2xcosx+4sinx-4=0 has exactly 3 solutions in the interval 0,nπ2,n∈N, then the roots of the equation x2+nx+(n-3)=0 belong to :

Choose an option

Show full solutionCorrect option: B
Correct answer
B(-∞,0)

Step-by-step explanation

Given: 2sin3x+sin2xcosx+4sinx-4=0

⇒2sin3x+2sinx·cos2x+4sinx-4=0

⇒2sin3x+2sinx·1-sin2x+4sinx-4=0

⇒2sin3x+2sinx-2sin3x+4sinx-4=0

⇒6sinx-4=0

⇒sinx=23

Now, for exactly three solution we get,

⇒n=5 (in the given interval)

So, x2+nx+n-3=0

⇒x2+5x+2=0

⇒x=-5±172

So, the required interval is (-∞,0).

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About this question

This is a previous-year question from JEE Main 2024, covering the Trigonometric Equations chapter of Mathematics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.