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Bisection method

  1. Feb 5, 2008 #1
    I need to solve this equaation:

    (x1cos(p)+y1sin(p))^2 * (x2sin(p)-y2cos(p)) = (x1sin(p)-y1cos(p)) * (x2cos(p)+y2sin(p))^2 ;

    x1,x2,y1,y2 are constants

    The equation would have 3 roots

    1 real and two imaginary

    I don't need imaginary roots.

    I am planning to use bisection method-----

    since i know the real root will lie between -2*pi and 2*pi

    What is your comment on this?
  2. jcsd
  3. Feb 5, 2008 #2


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    Without commenting on your method I see that the equation as stated doesn't make sense. You have unbalanced parentheses on both sides.
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