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Answer 1 · 2017-05-13T16:52:50

#x=pi/4+(npi)/2# where #ninZZ#

Question: #sec^2x+tan^2x=3tan^2x#

We can move the #tan^2x# to the right-hand side:
#sec^2x=2tan^2x#

Note that #secx=1/cosx# and #tanx=sinx/cosx#
So we have:
#1/cos^2x=(2sin^2x)/cos^2x#
#1/cos^2x-(2sin^2x)/cos^2x=0#
#(1-2sin^2x)/cos^2x=0#

Note the double angle identity for #cos(2x)=cos^2x-sin^2x=1-2sin^2x#
So we have:
#(cos^2x-sin^2x)/cos^2x=0#

#cos^2x/cos^2x-sin^2x/cos^2x=0#

#1-sin^2x/cos^2x=0#

#1-tan^2x=0#

#tan^2x=1#

#tanx=+-sqrt1#

#tanx=+-1#

Therefore,
#x=pi/4+(npi)/2# where #ninZZ#