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1995

JEE Main 1995 Mathematics question on Trigonometry

The general value of θ\theta satisfying the equation 2sin⁡2θ−3sin⁡θ−2=02\sin^2\theta - 3\sin\theta - 2 = 0, is
  1. A.
    nπ+(−1)nπ6n\pi + (-1)^n \dfrac{\pi}{6}
  2. B.
    nπ+(−1)nπ2n\pi + (-1)^n \dfrac{\pi}{2}
  3. C.
    nπ+(−1)n5π6n\pi + (-1)^n \dfrac{5\pi}{6}
  4. D.
    nπ+(−1)n7π6n\pi + (-1)^n \dfrac{7\pi}{6}
Show answer and solution

Correct option: D

Solution

Factoring: 2sin²θ - 3sinθ - 2 = (2sinθ + 1)(sinθ - 2) = 0.

Since sinθ = 2 is impossible, we need sinθ = -1/2.

The principal value where sinθ = -1/2 is θ = -π/6, but the standard form uses the equivalent angle: sin(7π/6) = -1/2, and 7π/6 lies in [0, 2π].

The general solution is θ = nπ + (-1)ⁿ(7π/6), since this formula with 7π/6 as the reference angle correctly captures sinθ = -1/2 for all integers n.

[1995]

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