So x=n*pi + theta. How do I determine what theta is? Theta<pi/2 but only by a little and it approaches pi/2 as n approaches infinity. I don't know how to find an exact solution though.
Correct.Since tan(n*pi)=0 it says that x=tan (theta)?
Veering off course. Make the substitution on both sides of the equality, and remember that a fixed point iteration based on tangent is not stable. You need to use atan.Theta=arctan (x). So tan(theta)=n*pi +arctan(x). I'm I veering off course? It kind of feels like it :P
It does if theta is in the principal domain of arctan. That was the point of creating that variable: to ensure that it does lie in the principal domain of arctan.Ok. Does arctan (tan ([tex]\theta))=\theta[/tex]?
Yes, you are. So, one step at a time.If it does, then arctan (n*pi + [tex]\theta[/tex])=([tex]\theta[/tex]). I think I'm going in circles.