How do you solve ##tan 4x = tan 2x##?

##{ pi/2 + 2kpi, k in Z }##
Recall
##tanx = sinx/cosx##
then,
##tan4x = (sin4x)/(cos4x) and tan2x = (sin2x)/(cos2x)##
##(sin4x)/(cos4x) = (sin2x)/(cos2x)## [Cross multiplying]
##sin4xcos2x = cos4xsin2x## [Let’s gather them all on one side]
##sin4xcos2x – cos4x.sin2x = 0##
Recall;
##sin[p – q] = sinpcosq – sinqcosp##
[it is effective to know summing up/extracting formulas for both ##sin## and ##cos##]
##sin4xcos2x – cos4x.sin2x = sin(4x – 2x) = sin2x##
##sin2x=0##
##sin2x = sinpi##
##2x=pi##
##x=pi/2##
Remember that the period of ##sin## function is ##2pi##
Solution set; ##{ pi/2 + 2kpi, k in Z }##