**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