The number of solutions of the pair of equations and in the interval is
- A.0
- B.1
- C.2
- D.4
Show answer and solution
Correct option: C
Solution
From the first equation: 2sin²θ - (1 - 2sin²θ) = 0 → 4sin²θ = 1 → sinθ = ±1/2.
From the second equation: 2(1 - sin²θ) - 3sinθ = 0 → 2sin²θ + 3sinθ - 2 = 0 → (2sinθ - 1)(sinθ + 2) = 0 → sinθ = 1/2 (since sinθ = -2 is impossible).
The common solution requires sinθ = 1/2, giving θ = π/6 and θ = 5π/6 in [0, 2π]. Both satisfy both equations, so there are 2 solutions.
From the second equation: 2(1 - sin²θ) - 3sinθ = 0 → 2sin²θ + 3sinθ - 2 = 0 → (2sinθ - 1)(sinθ + 2) = 0 → sinθ = 1/2 (since sinθ = -2 is impossible).
The common solution requires sinθ = 1/2, giving θ = π/6 and θ = 5π/6 in [0, 2π]. Both satisfy both equations, so there are 2 solutions.
[AIEEE 2007]