Trigonometric EquationHard
Question
The general value of θ satisfying the equation 2sin2 θ - 3 sin θ - 2 = 0, is
Options
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D.
Solution
Given, 2sin2 θ - 3sin θ - 2 = 0
⇒ (2sin θ + 1)(sin θ - 2) = 0
⇒ sinq = - 1/ 2 (neglecting sin θ = 2, as | sin θ | ≤ 1)
∴ q = nπ + (-1)n (7π / 6)
⇒ (2sin θ + 1)(sin θ - 2) = 0
⇒ sinq = - 1/ 2 (neglecting sin θ = 2, as | sin θ | ≤ 1)
∴ q = nπ + (-1)n (7π / 6)
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