SOLUTION: Solve to the nearest degree if needed, for 0≤x≤360° sin^2x=cosx 2sin^2x+cosx−cos^2x=0 Thank You!

Algebra ->  Trigonometry-basics -> SOLUTION: Solve to the nearest degree if needed, for 0≤x≤360° sin^2x=cosx 2sin^2x+cosx−cos^2x=0 Thank You!       Log On


   

Question 826960: Solve to the nearest degree if needed, for 0≤x≤360°
sin^2x=cosx
2sin^2x+cosx−cos^2x=0
Thank You!
Answer by lwsshak3(11628) About Me  (Show Source):
You can put this solution on YOUR website!
Solve to the nearest degree if needed, for 0≤x≤360°
***
For: sin^2x=cosx
1-cos^2x=cosx
cos^2x+cosx-1=0
solve for cosx by quadratic formula:
cosx+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
a=1, b=1, c=-1
ans:
cosx≈-1.618 (reject,(-1 cosx≈0.618
x≈51.83˚,308.17˚
..
For:2sin^2x+cosx−cos^2x=0
2(1-cos^x)+cosx-cos^2x=0
2-2cos^x)+cosx-cos^2x=0
3cos^2x-cosx-2=0
factor:
(3cosx+2)(cosx-1)=0
cosx=-2/3
x≈131.81˚,228.19˚
or
cosx=1
x=0