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August 30, 2014

August 30, 2014

Posted by **lindsay** on Tuesday, May 1, 2012 at 4:30pm.

- trigonometry -
**Reiny**, Tuesday, May 1, 2012 at 4:38pm4 tan^2 x + 12 sec x +1 =0

4sin^2 x/cos^2 x + 12/cosx + 1 = 0

times cos^2 x

4sin^2x + 12cosx + cos^2x = 0

4(1-cos^2 x) + 12cosx + cos^2 x = 0

-3cos^2 x + 12cosx + 4 = 0

3cos^2 x - 12cosx - 4 = 0

cosx = (12 ± √192)/6

= 4.309.. which is not possible, or cosx = -.3094..

so x must be in quads II or III

x = 180-71.977° or x = 180+71.977

x = 108.02° or 251.98°

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