SOLUTION: Hello, I need some help with a trig equation that wants me to find the solution for tan^2x+secx = 1 (0< x< 360)(<= less than or equal to). I got 0 degrees and 360 degrees, but I

Algebra ->  Trigonometry-basics -> SOLUTION: Hello, I need some help with a trig equation that wants me to find the solution for tan^2x+secx = 1 (0< x< 360)(<= less than or equal to). I got 0 degrees and 360 degrees, but I       Log On


   

Question 966066: Hello,
I need some help with a trig equation that wants me to find the solution for tan^2x+secx = 1 (0< x< 360)(<= less than or equal to). I got 0 degrees and 360 degrees, but I don't know the other answers.Any help would be greatly appreciated.Thanks!
K. Legters
Answer by lwsshak3(11628) About Me  (Show Source):
You can put this solution on YOUR website!
find the solution for tan^2x+secx = 1 [0< x< 360)
tan%5E2%28x%29%2Bsecx=1
%28sin%5E2%28x%29%2Fcos%5E2%28x%29%29%2B%281%2Fcosx%29=1
lcd: cos^2(x)
%28sin%5E2%28x%29%29%2B%28cosx%29=cos%5E2%28x%29
%281-cos%5E2%28x%29%29%2B%28cosx%29=cos%5E2%28x%29
2cos^2(x)-cosx-1=0
(2cosx+1)(cosx-1)=0
cosx=-1/2
x=120˚, 240˚(In quadrants II and III where cos < 0
or cosx=1
x=0