If tan⁡3θ+tan⁡θ=2tan⁡2θ, then θ is equal to n∈Z

tan⁡3θ+tan⁡θ=2tan⁡2θ,or  tan⁡3θ−tan⁡2θ=tan⁡2θ−tan⁡θor  sin⁡(3θ−2θ)cos⁡3θcos⁡2θ=sin⁡(2θ−θ)cos⁡2θcos⁡θor  sin⁡θ(2sin⁡θsin⁡2θ)=0or  sin⁡θ=0 or sin⁡2θ=0⇒θ=nπ or 2θ=nπ,n∈ZBut θ=nπ/2 is rejected because when n is odd, tan θ is not defined and when n is even, i.e., 2r, then θ=rπThen θ˙=nπ,n∈I is the only solution