Approximation of values from non-closed form equation.

AI Thread Summary
The equation 1 = cos²(b)[1-(c-b)²] is presented for approximation, with c > π, and the user seeks a method to find b to 5-6 digits of accuracy. It is noted that a closed-form solution for b is unlikely. Newton's method is suggested as a viable approach for iteration, using the formula x_{n+1} = x_n - f(x_n)/f'(x_n). The discussion highlights that for real solutions, the condition tan²(b) = -(c-b)² leads to the conclusion that c must equal b, specifically c = nπ. The conversation emphasizes the importance of iterative methods for approximating solutions in non-closed form equations.
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Hello everyone, I'm working on a problem and it turns out that this equation crops up:

1 = cos^{2}(b)[1-(c-b)^{2}]

where

c > \pi

Now I'm pretty sure you can't solve for b in closed form (at least I can't), so what I need to do is for some value of c, approximate the value of b to about 5-6 digits of accuracy. I just need tips to head in the right direction. Anything will be useful. Thank you!
 
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1 = cos^{2}(b)[1-(c-b)^{2}]
1-cos^{2}(b) = cos^{2}(b)[-(c-b)^{2}]
sin^{2}(b) = cos^{2}(b)[-(c-b)^{2}]
tan^{2}(b) = -(c-b)^{2}
For real solution, positive term = negative term is only possible if they are =0.
Hence the solution is : c=b=n\pi
 
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Oh okay, thanks JJacquelin, I didn't even think to do this.
 
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