1990

JEE Main 1990 Mathematics question on Trigonometry

In a △ABC\triangle ABC, angle AA is greater than angle BB. If the measures of angles AA and BB satisfy the equation 3sin⁡x−4sin⁡3x−k=0, 0<k<13\sin x - 4\sin^3 x - k = 0,\ 0 < k < 1, then the measure of ∠C\angle C is
  1. A.
    π3\dfrac{\pi}{3}
  2. B.
    π2\dfrac{\pi}{2}
  3. C.
    2π3\dfrac{2\pi}{3}
  4. D.
    5π6\dfrac{5\pi}{6}
Show answer and solution

Correct option: C

Solution

The equation 3sin x - 4sin³x = k means sin(3x) = k. So both A and B satisfy sin(3A) = sin(3B) = k.

Since A ≠ B, we need sin(3A) = sin(3B) with 3A ≠ 3B, which gives 3A + 3B = π (the supplementary case), so A + B = π/3.

Therefore C = π - (A + B) = π - π/3 = 2π/3.

[1990]

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