Question:
If 2sin2θ = 3cos θ, where 0 ≤ θ ≤ 2π, then find the value of θ.
Solution:
According to the question,
2sin2θ = 3cos θ
We know that,
sin2θ = 1 – cos2θ
Given that,
2 sin2 θ = 3 cos θ
2 – 2 cos2 θ = 3 cos θ
2 cos2 θ + 3 cos θ – 2= 0
(cos θ + 2)(2 cos θ – 1) = 0
Therefore,
cos θ = ½ = cos π/3
θ = π/3 or 2π – π/3
θ = π/3, 5π/3
Therefore, 2(1 – cos2θ) = 3cos θ
⇒ 2 – 2cos2θ = 3cos θ
⇒ 2cos2θ + 3cos θ – 2 = 0
⇒ 2cos2θ + 4cos θ – cos θ – 2 = 0
⇒ 2cos θ (cos θ+ 2) +1 (cos θ + 2) = 0
⇒ (2cos θ + 1)(cos θ + 2) = 0
Since, cos θ ∈ [-1,1] , for any value θ.
So, cos θ ≠ – 2
Therefore,
2 cos θ – 1 = 0
⇒ cos θ = ½
= π/3 or 2π – π/3
θ = π/3, 5π/3