Need help solving the following equation over the interval [0,2pi)

Question

tan2x-sec2x=2

Answer 1

 

tan2x-sec2x=2

tan=sin/cos and sec=1/cos

sin2x/cos2x-1/cos2x=2

(sin2x-1)/cos2x=2

sin2x=2sinxcosx and cos2x=cos^2x-sin^2x

(2sinxcosx-1)/(cos^2x-sin^2x)=2

sin^2x+cos^2x=1 so -1=-sin^2x-cos^2

substitute and then factor the denominator

(2sinxcosx-sin^2x-cos^2x)/(cosx-sinx)(cosx+sinx)=2

multiply numerator and denominator of the fraction by -1

(sin^2x-2sinxcosx+cos^2x)/(sinx-cosx)(sinx+cosx)=2

factor the numerator

(sinx-cosx)(sinx-cosx)/(sinx-cosx)(sinx+cosx)=2

simplify, factor out (sinx-cosx)

(sinx-cosx)/sinx+cosx)=2

sinx-cosx=2(sinx+cosx)

sinx-cosx=2sinx+2cosx

sinx-2sinx=cosx+2cosx

-sinx=3cosx

sinx=-3cosx

sinx/cosx=-3

tanx=-3

use arctan on a calculator or a table or...to find that...

x=-71.56505118º (round as needed) (tan[-71.565...º=-3)

check:tan2x-sec2x=tan(-143º)-sec(-143º)≈0.75-(-1.25)=0.75+1.25=2