How do I solve this for theta?

Question

$$\frac{\cos(\frac{\pi}{2}\cos{\theta})}{\sin{\theta}} = \frac{1}{\sqrt{2}} $$

How do I solve for $\theta$ here?

Answer 1

As I wrote in comments, only numerical methods would do the job.

Plot the function and locate approximate values of the solutions; for example, for $-10 \leq\theta \leq 10$, there solutions close to $-6,-4,1,2,7,8$.

Now, consider that you want to polish one of the roots; let us say the one close to $\theta_0=8$. Now, use Newton method which will update the solution according to $$\theta_{n+1}=\theta_n-\frac{f(\theta_n)}{f'(x_n)}$$ using $$f(\theta)=\cos \left(\frac{\pi}{2} \cos (\theta )\right) \csc (\theta )-\frac{1}{\sqrt{2}}$$ $$f'(\theta)=\frac{\pi}{2} \sin \left(\frac{\pi}{2} \cos (\theta )\right)-\cos \left(\frac{\pi}{2} \cos (\theta )\right) \cot (\theta ) \csc (\theta )$$ The successive iterates will then be $$\left( \begin{array}{cc} n & \theta_n \\ 0 & 8.00000 \\ 1 & 9.31385 \\ 2 & 8.52694 \\ 3 & 8.53536 \\ 4 & 8.53534 \end{array} \right)$$ which is the solution for six significant figures.

Similarly, using $\theta_0=-6$, you would get $$\left( \begin{array}{cc} n & \theta_n \\ 0 & -6.00000 \\ 1 & -5.39490 \\ 2 & -5.39375 \end{array} \right)$$ which is the solution for six significant figures.