The minimum value of 2sin²θ + 3cos²θ is?

Question: The minimum value of 2sin²θ + 3cos²θ is?

Solution:
Let,
p = 2sin²θ + 3cos²θ
⇒ p = 2sin²θ + 2cos²θ + cos²θ
⇒ p = 2(sin²θ + cos²θ) + cos²θ
⇒ p = 2 + cos²θ [since sin²θ + cos²θ = 1]

Therefore p will be the minimum when cosθ = 0.
So, the minimum value of p will 2.