How do you solve #sec^2xtan^2x+3sec^2x-2tan^2x=3# for #0<=x<=2pi#?

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Answer 1 · 2016-09-22T14:43:25

#0#, #pi# or #2pi#

#sec^2xtan^2x+3sec^2x-2tan^2x=3# can be converted in to an equation with #tanx# as #sec^2x=1+tan^2x#.

Using this identity #sec^2xtan^2x+3sec^2x-2tan^2x=3# becomes

#(1+tan^2x)tan^2x+3(1+tan^2x)-2tan^2x-3=0# or

#tan^2x+tan^4x+3+3tan^2x-2tan^2x-3=0# or

#tan^4x+2tan^2x=0# or

#tan^2x(tan^2x+2)=0#

As #tan^2x+2# cannot be zero, #tan^2x=0# or #tanx=0#

i.e. #x=npi#

and in the range #0<=x<=2pi#, the values are #0#, #pi# or #2pi#.