If 3+2isinθ1−2isinθ is real, then θ=

z=3+2isinθ1−2isinθ=(3+2isinθ)(1+2isinθ)(1−2isinθ)(1+2isinθ)=(3−4sin2θ)+8isinθ1+4sin2θ z is real ⇒ Imaginary part of z is 0⇒8sinθ=0⇒sinθ=0⇒θ=nπ,n∈Z.