How do you solve # Tan^2(x)-|Sec(x)|=1 #?

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Answer 1 · 2016-11-30T12:42:29

#x= 2kpi+-pi/3# and

#x= 2kpi+-2/3pi, k = 0, +-1, +-2, +-3, ..#

#|sec x| >=0#.

As #tan^2x = sec^2x-1 =|sec x|^2-1. #

the given equation becomes a quadratic in #|sec x|#.

#|sec x|^2-|sec x|-2=(|sec x|-2)(|xec x|+1)=0#.

So, #|sec x|=2 to sec x = +- 2#

#sec^(-1)(+-2)= pi/3 and 2/3pi#.. And so,

#x= 2kpi+-pi/3# and

#x= 2kpi+-2/3pi, k = 0, +-1, +-2, +-3, ..#