Question 1015621
tan A + tan2A + tan3A = 0

{{{(tanA + tan2A) + (tanA + tan2A)/(1-tanA*tan2A) = 0}}}

==> {{{(tanA + tan2A)*(1 + 1/(1-tanA*tan2A)) = 0}}}


==>{{{(tanA + tan2A)*((2 - tanA*tan2A)/(1-tanA*tan2A)) = 0}}}

==> tanA + tan2a = 0  OR {{{(2 - tanA*tan2A)/(1-tanA*tan2A) = 0}}}

The first case is equivalent to {{{(sinA*cos2A + sin2A*cosA)/(cosA*cos2A) = 0}}}, or equivalently, {{{(sin3A)/(cosA*cos2A) = 0}}}.
This happens only when 3A = 0, or A = 0 degree.  (Notice A = 0 does not make the denominator 0, so A = 0 is a valid solution.  It can be checked also by using the original equation that A = 0 is indeed a solution.)

The 2nd case {{{(2 - tanA*tan2A)/(1-tanA*tan2A) = 0}}} implies that 
2 - tanA*tan2A = 0.  (Notice that this effectively means that the denominator 1 - tanA*tan2A is not equal to 0.)

==> tanA*tan2A = 2
==> {{{tanA*((2tanA)/(1-(tanA)^2)) = 2}}} <==>{{{(2(tanA)^2)/(1-(tanA)^2) = 2}}}.
Now the only value of A that will make the denominator of the last equation 0 is 45 degrees, in which tanA = 1.  Thus we exclude A = 45 degrees from the solution and 
let {{{2(tanA)^2 = 2-2(tanA)^2}}}, or

{{{4(tanA)^2 = 2}}}, or {{{(tanA)^2 = 1/2}}}.

==> {{{tanA = 1/sqrt(2)}}} or {{{tanA = -1/sqrt(2)}}}.

The second instance is unacceptable, since it would situate A outside 0<_A<_90°.
Thus we have

{{{A = tan^-1(1/sqrt(2))}}}.

Therefore the solution set for the original equation is {0, {{{tan^-1(1/sqrt(2))}}} }