- #1
01suite
- 10
- 0
soo okay...i have this question that I am stuck on...
Solve for x,
cos 2x - cos^2x = 0, -180degrees<x<180degrees
i tried solving it...
cos 2x - cos^2x = 0
[2cos^2x -1]-cos^2x = 0
cos^2x -1 = 0 <--(?)
(cos x -1)(cos x +1) = 0
therefore... 1)cos x = -1 , 2)cos x = 1
so...
1)cos x = -1
x = cos^-1 (-1)
x= 180
2) cos x = 1
x = 0
but i have tried on my calculator...and -180 degrees can be one of the answers too...i don't get how to get -180
SOMEONE HELPP...and reply FASTT please and Thank yOU in advance
Solve for x,
cos 2x - cos^2x = 0, -180degrees<x<180degrees
i tried solving it...
cos 2x - cos^2x = 0
[2cos^2x -1]-cos^2x = 0
cos^2x -1 = 0 <--(?)
(cos x -1)(cos x +1) = 0
therefore... 1)cos x = -1 , 2)cos x = 1
so...
1)cos x = -1
x = cos^-1 (-1)
x= 180
2) cos x = 1
x = 0
but i have tried on my calculator...and -180 degrees can be one of the answers too...i don't get how to get -180
SOMEONE HELPP...and reply FASTT please and Thank yOU in advance