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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 }##

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