Question

If 2sin^2 θ+ 3sinθ = 0, then permissible values of cosθ are ...​

Answer 1

Answer:

2sin²θ + 3sinθ = 0

∴ sinθ (2sinθ + 3) = 0

∴ sinθ = 0 or sinθ = -3/2

Since, - 1 sinθ ≤ 1,

sinθ = 0

∴√1 - cos²θ = 0 [ ∵sin²θ = 1 - cos²θ ]

∴ 1 - cos²θ = 0

∴ cos²θ = 1

∴cosθ ± 1 [ ∵1 ≤ cosθ ≤ 1 ]