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

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.

  • trigonometry - , Tuesday, May 1, 2012 at 4:38pm

    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(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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