tan(A+B)=(tanA+tanB)/(1-tanAtanB).
Therefore tan(A+B+C)=(tan(A+B)+tanC)/(1-tan(A+B)tanC)=
((tanA+tanB)/(1-tanAtanB)+tanC)/(1-tan(A+B)tanC)=
(tanA+tanB+tanC-tanAtanBtanC)/(1-tan(A+B)tanC).
The numerator is zero when tanA+tanB+tanC=tanAtanBtanC. This implies that tan(A+B+C)=0, therefore A+B+C=nπ because tan(nπ)=0 for all integer n. C=nπ-A-B. If A=x and B=2x, then C=nπ-3x=7x. Therefore 10x=nπ, x=nπ/10 is the solution. Generally, the factor nπ/(A+B+C) will be part of the solution.
However, when n=5, x=π/2 and tan(π/2) is undefined so we have to avoid multiples of π/2. x=0, π/10, π/5, 3π/10, 2π/5, 3π/5, 7π/10, 4π/5, 9π/10, 11π/10, etc. n cannot be a multiple of 5, in this case.