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Question #e4bf6

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Mar 12, 2017

#x=0, pi/2, pi, (3pi)/2#
#x=(3 pi)/8, (7pi)/8, (11pi)/8, (15pi)/8 #

Explanation:

#(sec2x)^2 -1#equals #(tan 2x)^2#. The given equation thus becomes tan 2x (tan 2x +1)=0

It is thus tan 2x=0 or tan 2x= -1

Tan 2x=0 gives #2x=0, pi, 2pi,3pi# that is #x=0, pi/2, pi, (3pi)/2#

Tan 2x=-1, gives #2x= (3pi)/4, (7pi)/4, (11pi)/4, (15pi)/4, # so that

#x=(3 pi)/8, (7pi)/8, (11pi)/8, (15pi)/8 #

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