Question

If 2 sin²θ + 3 sin θ = 0, find the permissible values of cos θ.

Answer 1

2 sin²θ + 3 sinθ = 0

∴ sin θ (2 sin θ + 3) = 0

∴ sin θ = 0 or 2 sin θ + 3 = 0

∴ sin θ = 0 or sin θ = `- 3/2`

Since – 1 ≤ sin θ ≤ 1, for all θ ∈ R.

∴ sin θ ≠ `- 3/2`

∴ sin θ = 0

∴ θ = nπ, n ∈ Z.

∴ `sqrt(1 - cos^2θ) = 0   ...[(sin^2θ = 1  –  cos^2θ),(∵ sin θ = sqrt(1 - cos^2θ))]`

∴ 1 – cos²θ = 0

∴ cos²θ = 1

∴ cos θ = ± 1                 ...[∵ – 1 ≤ cosθ ≤ 1]

Answer 2

We know that sin²θ = 1 − cos²θ

∵ `sin θ = sqrt(1 - cos^2θ)`

Given equation 2 sin²θ + 3 sin θ = 0 becomes

`2(1 − cos^2θ) + 3 (sqrt(1 - cos^2θ)) = 0`

∴ 2(1 − cos²θ) + 3 (1 - cos²θ) = 0

∴ 5(1 − cos²θ) = 0

∴ 1 − cos²θ = 0

∴ cos²θ = 1

∴ cos θ = ± 1            ...[∵ – 1 ≤ cosθ ≤ 1]

∴ The permissible value of cos θ is ± 1.