Question

If 2sin²θ = 3cosθ, where 0 ≤ θ ≤ 2π, then find the value of θ.

Answer 1

2sin²θ = 3cosθ

We know that,

sin²θ = 1 – cos²θ

Given that,

2sin²θ = 3 cosθ

2 – 2cos²θ = 3cosθ

2cos²θ + 3cosθ – 2 = 0

(cosθ + 2)(2cosθ – 1) = 0

Therefore,

cosθ = `1/2 = cos  pi/3`

θ = `pi/3` or `2π  –  pi/3`

θ = `pi/3, (5pi)/3`

Therefore, 2(1 – cos²θ) = 3cosθ

⇒ 2 – 2cos²θ = 3cosθ

⇒ 2cos²θ + 3cosθ – 2 = 0

⇒ 2cos²θ + 4cosθ – cosθ – 2 = 0

⇒ 2cosθ(cosθ + 2) + 1(cosθ + 2) = 0

⇒ (2cosθ + 1)(cosθ + 2) = 0

Since, cosθ ∈ [–1, 1], for any value θ.

So, cosθ ≠ –2

Therefore,

2cosθ – 1 = 0

⇒ cosθ = `1/2`

= `pi/3` or `2π  –  pi/3`

θ = `π/3, (5pi)/3`