What is the general solution of the equation #sec^2x-tanx=4#?

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Answer 1 · 2017-09-12T15:06:09

#x=nxx180^@+66.53^@# or #nxx180^@-52.49^@#, where #n# is an integer.

#sec^2x-tanx=4# means

#1+tan^2x-tanx=4#

or #tan^2x-tanx-3=0#

and using quadratic formula #tanx=(-(-1)+-sqrt((-1)^2-4*1*(-3)))/2#

= #(1+-sqrt13)/2#

= #(1+-3.6056)/2#

i.e. #tanx=2.3028# or #-1.3028#

i.e. #tanx=tan66.53^@# or #tan(-52.49^@)#

Hence #x=nxx180^@+66.53^@# or #nxx180^@-52.49^@#, where #n# is an integer.