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Math Help - Trigo Angles

  1. #1
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    Trigo Angles

    Solve the equation 6x^3 + 11x^2 - 3x - 2 = 0
    Hence, find the values of angle [D] between 0 and 180 degrees, which satisfy the equation 6 tan[D]^2 + 11 tan [D] - 3 = 2 cot [D]

    I have solved the cubic equation by factor theorem - its answers are
    -2, -1/3, 1/2

    I can't seem to find the values of angle [D], since the new equation doesn't exactly tally with the cubic one.
    Please assist.

    Thank you.
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  2. #2
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    Quote Originally Posted by Drdj View Post
    Solve the equation 6x^3 + 11x^2 - 3x - 2 = 0
    Hence, find the values of angle [D] between 0 and 180 degrees, which satisfy the equation 6 tan[D]^2 + 11 tan [D] - 3 = 2 cot [D]

    I have solved the cubic equation by factor theorem - its answers are
    -2, -1/3, 1/2

    I can't seem to find the values of angle [D], since the new equation doesn't exactly tally with the cubic one.
    Please assist.

    Thank you.
    Yes it does...

    6\tan^2{D} + 11\tan{D} - 3 = 2\cot{D}

    Multiply everything by \tan{D}

    6\tan^3{D} + 11\tan^2{D} - 3\tan{D} = 2

    6\tan^3{D} + 11\tan^2{D} - 3\tan{D} - 2 = 0.


    Now by letting x = \tan{D}, you have the same cubic equation...
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  3. #3
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    Thanks

    Thank you so much !! Never would have thought of that !!
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  4. #4
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    Thanks Prove It, I learnt something too!!
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