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Math Help - Trig problem, ill paypal you 5 dollars if its right.

  1. #1
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    Exclamation Need help, dont know how to solve, if you do, ill paypal you 5 dollars if its right.

    cos x csc^2 x + 3 cos x = 7 cos x

    Thanks,
    Mike
    Last edited by MathGuru; September 15th 2005 at 03:38 PM.
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  2. #2
    MHF Contributor
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    cos x csc^2 x + 3 cos x = 7 cos x

    Rewriting that,
    cosXcsc^2(X) +3cosX = 7cosX
    cosXcsc^2(X) +3cosX -7cosX = 0
    cosXcsc^2(X) -4cosX = 0
    cosX[csc^2(X) -4] = 0

    cosX = 0
    X = arccos(0) = 90 or 270 degrees

    csc^2(X) -4 = 0
    csc^2(X) = 4
    cscX = +,-2

    When cscX = 2,
    cscX = 1/sinX
    So, 1/sinX = 2
    sinX = 1/2 ---positive
    Sine is positive in the 1st and 2nd quadrants, so,
    X = arcsin(1/2)
    X = 30 deg --in the 1st quadrant
    X = 180-30 = 150 deg ---in the 2nd quarter

    When cscX = -2,
    1/sinX = -2
    sinX = -1/2 ---negative
    Sine is negative in the 3rd and 4th quadrants, so,
    X = arcsin(-1/2)
    X = 180 +30 = 210 deg --in the 3rd quadrant
    X = 360 -30 = 330 deg ---in the 4th quarter

    Therefore, if X is from 0 to 360 degrees only,
    X = 30, 90, 150, 210, 270, or 330 degrees -------answer.

    Check those against the original equation and you'd find them all correct.

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  3. #3
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    Thank you so much ticbol, your a savior
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