1sumit ur correct
ans is A
general solution for sin θ = k is θ = nπ + ( - 1 ) n α
u got sin
2θ = 34
sin θ = ± 32
therefore α = ± π3
therefore general equation is
θ = nπ + ( -1 ) n (±π3 )
which is ( a )
7 Answers
62two ways either try to express in terms of r and theta in the numerator and denominator
Or take this whole thing to be equal to k.i (where k is a constant)
1i m getting this expression as 3+8isin θ-4sin2θ1+4sin2θ
so 3-4sin2θ=0
sin2θ=3/4
which is satisfying only option d.
am i correct?????????
1
29in that question they asked abt the real part...not the imaginary part