tan(4x)=tan(2x)=2tan(2x)/(1-tan2(2x)).
Let t=tan(2x): t=2t/(1-t2), t(1-2/(1-t2))=0, so t=0 is one solution.
1-t2-2=0, -t2-1=0, t2=-1 has no real solution. Therefore tan(2x)=0 is the only real solution.
2x=nπ, x=nπ/2, where n is an integer. The lowest multiple is 2x, so x=π/2, 3π/2, 5π/2, ... are valid solutions for x even though tan(x) would be undefined. So x=0 and π/2 for the given interval.