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Math Help - Roots on an equation

  1. #1
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    Roots on an equation

    The equation 2-cos^2\theta=Acos2\theta where A is a constant, has a root \theta=30^o. Find all the other roots such that \theta is between 0^o and 360^o
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  2. #2
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    Quote Originally Posted by Punch View Post
    The equation 2-cos^2\theta=Acos2\theta where A is a constant, has a root \theta=30^o. Find all the other roots such that \theta is between 0^o and 360^o
    Did you manage to find A?
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  3. #3
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    Yes i did, A = 1.549
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  4. #4
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    Quote Originally Posted by Punch View Post
    Yes i did, A = 1.549
    ok, then use the identity cos 2a = 2 cos^2 a -1
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  5. #5
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    Quote Originally Posted by mathaddict View Post
    ok, then use the identity cos 2a = 2 cos^2 a -1
    I guess the idea is to find the value of the constant A? How do I find the other roots after that? confused
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  6. #6
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    Quote Originally Posted by Punch View Post
    I guess the idea is to find the value of the constant A? How do I find the other roots after that? confused
    In your original equation,

    2-cos^2 b = A cos 2b

    2- cos^2 b = A (2cos ^2 b-1)

    2- cos^2 b = 2A cos^2 b-A

    Rearrange this: 2A cos^2 b+ cos^2 b- A- 2=0

    (2A+1) cos^2 b - A -2 =0

    Now you have got your A, plug it in here.
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  7. #7
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    Quote Originally Posted by mathaddict View Post
    In your original equation,

    2-cos^2 b = A cos 2b

    2- cos^2 b = A (2cos ^2 b-1)

    2- cos^2 b = 2A cos^2 b-A

    Rearrange this: 2A cos^2 b+ cos^2 b- A- 2=0

    (2A+1) cos^2 b - A -2 =0

    Now you have got your A, plug it in here.
    sorry, but after substituting the value of A in, what do I do?
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  8. #8
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    Quote Originally Posted by Punch View Post
    sorry, but after substituting the value of A in, what do I do?
    It seems like you are unsure how to solve s trigonometric equation. Review your class notes which should have covered that to get some basic understanding on this before you get back to this question.

    Anyways, in case if it's the square of the cosine which is confusing you,

    \cos b =\pm \sqrt{\frac{A+2}{2A+1}}
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