Question Number 14330 by myintkhaing last updated on 01/Jun/17
$${tan}\theta+{tan}\mathrm{2}\theta={tan}\mathrm{3}\theta \\ $$$$\frac{{tan}\theta+{tan}\mathrm{2}\theta}{\mathrm{1}−{tan}\theta{tan}\mathrm{2}\theta}\left(\mathrm{1}−{tan}\theta{tan}\mathrm{2}\theta\right)={tan}\mathrm{3}\theta \\ $$$${tan}\mathrm{3}\theta\left(\mathrm{1}−{tan}\theta{tan}\mathrm{2}\theta\right)={tan}\mathrm{3}\theta \\ $$$${tan}\theta{tan}\mathrm{2}\theta{tan}\mathrm{3}\theta\:\mathrm{0} \\ $$$$\mathrm{3}\theta={k}\pi,\:{where}\:{k}=\mathrm{0},\mathrm{1},\mathrm{2},\mathrm{3},... \\ $$$$\theta=\frac{{k}\pi}{\mathrm{3}} \\ $$$${tan}\theta=\mathrm{0}\:\:{or}\:\:{tan}\mathrm{2}\theta=\mathrm{0}\:{or}\:{tan}\:\mathrm{3}\theta\:=\:\mathrm{0} \\ $$$$\theta={m}\pi,\:{where}\:{m}=\mathrm{0},\mathrm{1},\mathrm{2},\mathrm{3},... \\ $$$$\mathrm{2}\theta={n}\pi,\:{where}\:{n}=\mathrm{0},\mathrm{1},\mathrm{2},\mathrm{3},... \\ $$$$\theta=\frac{{n}\pi}{\mathrm{2}}\:{but}\:\theta\notin\frac{\left({p}+\mathrm{1}\right)\pi}{\mathrm{2}},\:{p}=\mathrm{0},\mathrm{1},\mathrm{2},\mathrm{3},... \\ $$$${The}\:{exhaustive}\:{set}\:=\left\{\theta/\theta=\frac{{k}\pi}{\mathrm{3}}\:{or}\:\theta={m}\pi\right\} \\ $$
Commented by Tinkutara last updated on 30/May/17
$$\mathrm{Thanks}! \\ $$