Let
z=1−isinθ1+2isinθ
⇒z=1−isinθ1+2isinθ×1+isinθ1+isinθ
⇒z=1+sin2θ1−2sin2θ+3isinθ
⇒z=(1+sin2θ1−2sin2θ)+i(1+sin2θ3sinθ)
Since, z is a purely imaginary number, so real part must be zero, hence
1+sin2θ1−2sin2θ=0
⇒1−2sin2θ=0
⇒cos2θ=0
⇒θ=4π,43π,45π,47π, for θ∈(0,2π)
∴Sum of all values =416π=4π