What I'm trying to calculate is the following $$\int_0^{2\pi}dz_1\sum_{j\text{ s.t. } z_2^{(j)}=\arccos(\lambda-\cos(z1))\text{ or }\\\text{ written another way}\\0\leq\arccos(\lambda-\cos(z1))\leq\frac{2\pi}{B}}\frac{1}{\sqrt{1-\left(\lambda-\cos(z_1)\right)^2}}.$$

I want to find $z_2^{(j)}$ such that $$F(z_2^{(j)})=0,$$ where there are $j$ roots to this equation. I want to find $j$ such that $z_2^{(j)}\in[0,\frac{2\pi}{B}]$ ($B$ is a constant) and $$F(z_2)=\cos(z_2)-(\lambda-\cos(z_1))$$ and where $\lambda$ is a parameter ($\lambda\in(-2,2)$) which I want to plot over later, but for now I want to leave it as a constant

MY ANSWER: In my example I set $B=2$, but it would be nice to have it for a general $B$. I think $j=1$ since re-arranging we have $$\cos(z_2)=\lambda-\cos(z_1)$$ and since $\cos(z_2)$ in bijective in this interval there is only one solution. But this doesn't restrict $\cos(z_2)$ since my $\lambda\in(-2,2)$, this will cause the sqrt to be complex which by $$\cos(z_2)=\lambda-\cos(z_1)$$ cannot happen since $|\cos(z_2)|\leq1$.

How would one go about solving such a problem analytically? If not can someone possibly point me in the direction of some python document where a similar problem is tackled? I have tried to implement it in python using an if statement, but since I want to integrate it with respect to the argument that is being restricted it's quite difficult... Any help would be greatly appreciated!


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