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February 1, 2015

February 1, 2015

Posted by **Wendy** on Friday, September 16, 2011 at 4:44pm.

interval [0,2pi)

How do I solve this? Help would be greatly appreciated!

- Trig -
**Damon**, Friday, September 16, 2011 at 5:30pm2(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

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