The set of values of x satisfying the equation sin⁡3α=4sin⁡αsin⁡(x+α)sin⁡(x−α) is

We have sin⁡3α=4sin⁡αsin⁡(x+α)sin⁡(x−α)or  3sin⁡α−4sin3⁡α=4sin⁡αsin2⁡x−4sin3⁡αor  3sin⁡α=4sin⁡αsin2⁡xIf sin⁡α≠0,sin2⁡x=3/4=(3/2)2=sin2⁡(π/3)Therefore x=nπ±π/3,∀n∈ZIf sin⁡α=0, i.e., α=nπ,  equation becomes an identity.