Variable Separation - Find $\theta$ in $\frac{\cos\left(\frac{\pi}2\cos\theta\right)}{\sin\theta} = \frac{0.8912r}{60I}$ Could anyone kindly help me out to obtain an expression for θ in the given equation - 
$$\frac{\cos\left(\frac{\pi}2\cos\theta\right)}{\sin\theta} = \frac{0.8912r}{60I}$$
I can't separate $\theta$ from this expression. 
Thanks in advance!  
 A: This is a transcendental equation and there is no analytical solution (remember that this is already the case for $x=\cos(x)$) and numercial methods should be required for solving $$\cos \left(\frac{\pi}{2}   \cos (x)\right) \csc (x)=a$$ If you look at the plot of function $$f(x)=\cos \left(\frac{\pi}{2}   \cos (x)\right) \csc (x)$$ for positive values of $x$, you will notice that it goes through a maximum at $x=\frac \pi 2$ for which $f(\frac \pi 2)=1$ and a minimum at $x=\frac {3\pi} 2$ for which $f(\frac{3 \pi} 2)=-1$. 
So solutions exist if $-1\leq a \leq 1$.
For a given value of $a$, locate, using graphics, the root you want and start using Newton method. 
If you are concerned by the first positive root and a rather small $a$, you could approximate the function using Taylor expansion around $x=0$; this gives $$f(x)=\frac{\pi  x}{4}+O\left(x^3\right)$$ and then an approximate solution is $$x=\frac{4 a}{\pi }$$ This could be your first guess.
Similarly, around $x=\frac \pi 2$, composing Taylor series, you could obtain $$f(x)=1+\left(\frac{1}{2}-\frac{\pi ^2}{8}\right) \left(x-\frac{\pi
   }{2}\right)^2+O\left(\left(x-\frac{\pi }{2}\right)^3\right)$$ Solving the quadratic would give you an estimate for larger values of $a$.
