Question
2sin^2(x) = 2 + cos(x)
interval [0,2pi)
How do I solve this? Help would be greatly appreciated!
interval [0,2pi)
How do I solve this? Help would be greatly appreciated!
Answers
Damon
2(1-cos^2 x) = 2 + cos x
2 - 2 cos^2 x = 2 + cos x
2 cos^2 x + cos x = 0
cos x (2 cos x + 1 ) = 0
cos x = 0 when x = pi/2 or 90 degrees or 3 pi/2 or 270 deg
or
cos x = -1/2 at x = 2pi/3 or 120 degrees or at 4 pi/3 or 240 degrees
2 - 2 cos^2 x = 2 + cos x
2 cos^2 x + cos x = 0
cos x (2 cos x + 1 ) = 0
cos x = 0 when x = pi/2 or 90 degrees or 3 pi/2 or 270 deg
or
cos x = -1/2 at x = 2pi/3 or 120 degrees or at 4 pi/3 or 240 degrees