The correct option is B 2π3
Given :
θ∈(−π2,π) and 3 + 2isinθ1 − 2ιsinθ is purely imaginary
Let us assume Z=3 + 2isinθ1 − 2isinθ,
Multiplying and dividing Z with 1+2isinθ, we will get:
Z=(3 + 2isinθ) (1 + 2isinθ)12 − (2isinθ)2
⇒Z=3 + 6isinθ + 2isinθ + 4i2sin2θ1 − 4i2sin2θ
⇒Z=3 + 8isinθ − 4sin2θ1 + 4sin2θ
⇒Z=3 − 4sin2θ1 + 4sin2θ +i(8sinθ1 + 4sin2θ)
Since Z is purely imaginary,
∴3 − 4sin2θ1 + 4sin2θ = 0
⇒3−4sin2θ=0
⇒sin2θ=34
⇒sinθ=√34=±√32
⇒θ=−π3,π3,2π3 [∵θ∈(−π2,π)]
∴ A={−π3,π3,2π3}
Hence the sum of all the values of A =−π3+π3+2π3=2π3