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 }##