Question
Solve for the following equation to the nearest degree if needed for 0 less than equal to x less than equal to 360.
sin^2x=cosx
2sin^2x+cosx-cos^2x=0
sin^2x=cosx
2sin^2x+cosx-cos^2x=0
Answers
Steve
sin^2x = cosx
1-cos^2x = cosx
cos^2x + cosx - 1 = 0
cosx = 1/2 (√5 - 1)
x = 2kπ ± 0.90456
2sin^2x + cosx - cos^2x = 0
1-2cos^2x + cosx - cos^2x = 0
3cos^2x - cosx - 1 = 0
cosx = 1/6 (1±√13)
x = kπ or 2kπ ± 2.3005
1-cos^2x = cosx
cos^2x + cosx - 1 = 0
cosx = 1/2 (√5 - 1)
x = 2kπ ± 0.90456
2sin^2x + cosx - cos^2x = 0
1-2cos^2x + cosx - cos^2x = 0
3cos^2x - cosx - 1 = 0
cosx = 1/6 (1±√13)
x = kπ or 2kπ ± 2.3005