JEE Main 2025 · 3 April · evening · Single correct · +4 / −1

The number of solutions of the equation
(4-√3) sin x-2 √3 cos ² x=-4/(1+√3), x ∈[-2 π, (5 π)/2] is

  1. (a) 4
  2. (b) 3
  3. (c) 6
  4. (d) 5
Answer and solution

Answer (d) 5

Solution

Maths sir · 3 steps

Turn cos² x into 1-sin² x, and you get a quadratic in sin x that factorises neatly.

  1. Make it one function. Put cos² x = 1 - sin² x. Rationalise the right side: -4/(1+√3) = -4(√3-1)/2 = 2 - 2√3. The -2√3 terms on both sides cancel.

    2√3 sin² x + (4-√3)sin x - 2 = 0 ⇒ (2 sin x - 1)(√3 sin x + 2) = 0

  2. Keep valid root. The roots are sin x = 1/2 and sin x = -2/√3 ≈ -1.15. The second is less than -1, so it is impossible. Only sin x = 1/2 survives.
    Part of the intervalSolutions of sin x = 1/2
    [-2π, 0)-11π/6, -7π/6
    [0, 2π)π/6, 5π/6
    [2π, 5π/2]13π/6 only, since 17π/6 > 15π/6
  3. So. There are 2 + 2 + 1 = 5 solutions, which is option (d).

The idea to keep. With a mix of sin and cos², rewrite everything in sin x, solve the quadratic, and throw out any root outside [-1, 1]. Then count period by period, checking the last partial period carefully.

Answer from the NTA final answer key. Solution by PrepWiser.

Study Trigonometric Equations

Asked about 50 times in JEE Main since 2020. This question is from 2025.