Let S = θ ∈ [ 0 , 2 π ) : tan πcosθ + tan πsinθ = 0 , then ∑ θ ∈ S sin 2 θ + π 4 is equal to

Step-by-step solution

Given: S = θ ∈ [ 0 , 2 π ) :   tan πcosθ + tan πsinθ = 0 So, tan πcosθ + tan πsinθ = 0 ⇒ tan πcosθ = - tan πsinθ ⇒ tan πcosθ = tan - πsinθ ⇒ πcosθ = n π - πsinθ ;   n ∈ Z ⇒ sinθ + cosθ = n Now, - 2 ≤ sinθ + cosθ ≤ 2 ⇒ - 2 ≤ n ≤ 2 But n ∈ Z , so n = - 1 ,   0 ,   1 So, θ ∈ 0 , &#960