The set of values of satisfying the inequation , where , is
- A.
- B.
- C.
- D.None of the above
Show answer and solution
Correct option: A
Solution
Factor: 2sin²θ - 5sinθ + 2 = (2sinθ - 1)(sinθ - 2) > 0.
Since sinθ - 2 < 0 always, dividing by it flips the inequality: 2sinθ - 1 < 0, i.e., sinθ < 1/2.
In (0, 2π), sinθ < 1/2 holds when θ ∈ (0, π/6) ∪ (5π/6, 2π) — the endpoints θ = π/6 and 5π/6 give sinθ = 1/2, making the expression zero (not > 0), so they are excluded, giving open intervals.
Since sinθ - 2 < 0 always, dividing by it flips the inequality: 2sinθ - 1 < 0, i.e., sinθ < 1/2.
In (0, 2π), sinθ < 1/2 holds when θ ∈ (0, π/6) ∪ (5π/6, 2π) — the endpoints θ = π/6 and 5π/6 give sinθ = 1/2, making the expression zero (not > 0), so they are excluded, giving open intervals.
[AIEEE 2006]