Solving Equation with Bisection Method - Real Root Between -2*pi and 2*pi

[jrm2002]

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?

 

[mathman]

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