JEE Main 2024 Mathematics question
Not in JEE Main syllabus
If 2sin³x + sin2xcosx + 4sinx – 4 = 0 has exactly 3 solutions in the interval [0, nπ/2], n ∈ ℕ, then the roots of the equation x² + nx + (n – 3) = 0 belong to:
Answer: (A)
Solution
2sin³x + 2sin x cos²x + 4sin x − 4 = 0 gives 2sin x(sin²x + cos²x) + 4 sin x − 4 = 0, i.e. 6 sin x = 4, sin x = 2/3. It has 2 solutions in [0,
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