trigonometry
posted by lindsay .
solve the equation: 4 tan^2 x + 12 sec x +1 =0, for 0 degrees is less than or equal to x is less than or equal to 360 degrees.

4 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(1cos^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 = 18071.977° or x = 180+71.977
x = 108.02° or 251.98°
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