Let, tanx=u, tan2x=v.
Then, tan3x=tan(x+2x)=(tanx+tan2x)/(1-tanxtan2x), i.e.,
tan3x=(u+v)/(1-uv).
:. tanx+tan2x+tan3x=0,
rArr u+v+(u+v)/(1-uv)=0,
rArr (u+v){1+1/(1-uv)}=0,
rArr{(u+v)(2-uv)}/(1-uv)=0,
rArr (u+v)(2-uv)=0,
rArr v=-u, or, uv=2.
v=-u rArr tan2x=-tanx=tan(-x).
Recall that, tantheta=tanalpha rArr theta=kpi+alpha, k in ZZ.
"Case 1 : "v=-u
:. tan2x=tan(-x)
rArr 2x=kpi+(-x), or, 3x=kpi, i.e., x=kpi/3, k in ZZ.
"Case 2 : "uv=2.
:. tanx*tan2x=2,
Using tan2x=(2tanx)/(1-tan^2x), we have, then
(2tan^2x)/(1-tan^2x)=2, or, tan^2x=1-tan^2x, i.e.,
tan^2x=1/2 rArr tanx=+-1/sqrt2,
rArr x=kpi+arc tan(+-1/sqrt2), k in ZZ.
Altogether, the Solution Set is given by,
{kpi/3}uu{kpi+-arc tan(1/sqrt2)}, k in ZZ.
