..
#sec^2(4x)-sec(4x)=0#
#sec(4x)(sec(4x)-1)=0#
#sec(4x)=0#
This means #1/cos(4x)=0# which means #cos4x=oo#
#cos# values can only be between #-1# and #1#. As such, this answer is not valid.
#sec(4x)-1=0#
#sec(4x)=1#
#1/cos(4x)=1#
#cos4x=1#
#4x=0, 2pi, 4pi, 6pi,.......#
#4x=2kpi#
#x=(2kpi)/4=(kpi)/2#