Question #75174

1 Answer
Dec 19, 2017

#x=(kpi)/2#

Explanation:

..

#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#