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#.
