Let f ( θ ) = 3 sin 4 3 π 2 - θ + sin 4 ( 3 π + θ ) - 2 1 - sin 2 2 θ and S = θ ∈ [ 0 , π ] : f ' ( θ ) = - 3 2 . If 4 β = ∑ θ ∈ S θ then f ( β ) is equal to

Step-by-step solution

Given, f ( θ ) = 3 sin 4 3 π 2 - θ + sin 4 ( 3 π + θ ) - 2 1 - sin 2 2 θ and S = θ ∈ [ 0 , π ] : f ' ( θ ) = - 3 2 f ( θ ) = 3 sin 4 3 π 2 - θ + sin 4 ( 3 π + θ ) - 2 1 - sin 2 2 θ ⇒ f ( θ ) = 3 cos 4 θ + sin 4 θ - 2   cos 2 2 θ ⇒ f ( θ ) = 3 1 - 1 2 sin 2 2 θ - 2   cos 2 2 θ ⇒ f ( θ ) = 3 - 3 2 sin 2 2 θ - 2   cos 2 θ ⇒ f θ = 3 2 - 1 2 co