We have, `(tan3x-tan2x)/(1+tan3x.tan2x)=1 rArr tan(3x-2x)=1 rArr tanx=1`
`rArr tanx=tanx/4 rArr x=npi+pi/4,n in I` (using `tantheta=tanalpha ltimplies theta= npi+alpha)`
But for this value of x, `tan 2x` is not defined.
Hence the solution set for `x` is `phi`.
