The minimum value of 2sin2θ + 3cos2θ is ?

Answer: Option C

Let x = 2sin²θ + 3cos²θ
⇒ x = 2sin²θ + 2cos²θ + cos²θ
⇒ x = 2(sin²θ + cos²θ) + cos²θ
⇒ x = 2 + cos²θ     [since sin²θ + cos²θ = 1]
Therefore x will be the minimum when cosθ = 0. i.e. minimum value of x will 2

Alternative Solution:
2sin²θ + 3cos²θ
Minimum value is 2,
[If x sin²θ + y cos²θ, If x > y, then x will be always maximum value and y is minimum if y > x, vice versa will happen]