The real value of 'θ' for which the expression 3+2isinθ1−2isinθ is purely imaginary is
nπ, n∈Z
nπ±π2, n∈Z
nπ±π3, n∈Z
nπ±π6, n∈Z
3+2isinθ1−2isinθ is purely imaginary
⇒ Real part is zero
∴3−4sin2θ1+(2sinθ)2=0
⇒sin2θ=34
⇒sin2θ=±32
∴=±sinπ3
∴ Solution is θ=nπ±π3,n∈z