This is a very interesting question that I came across and have never solved any question of this sort. How do I find the largest open set on which $\frac{1}{cosz-2i}$ is analytic? Do I find the set in which $\cos z-2i\neq0$? Any answers will be much appreciated.
Note that $$\cos z=\frac{e^{iz}+e^{-iz}}{2}$$ where $z\in\mathbb{C}$
On
This is a ratio of two analytic functions, so it is analytic wherever the denominator is nonzero. (The numerator is constant, so in fact it is analytic exactly on the complement of the zero set of the denominator.) So, like you say, it is analytic precisely on the complement of the set where $\cos z - 2 i = 0$.
First, recall that $$ \cos(x+yi) = \cosh(y)\cos(x)-i\sinh(y)\sin(x). $$ If you haven't seen this before, it can be verified with a simple calculation.
So to have $\cos(x+yi)=2i$ we need to have $\cosh(y)\cos(x)=0$. Since $\cosh(y)>0$ for all $y$, we must have $\cos(x)=0$. Thus $x=\frac\pi2+\pi n$ for some $n\in\mathbb Z$ and therefore $\sin(x)=(-1)^n$.
Now we must have $\sinh(y)=-(-1)^n2$. Therefore $\cos(x+yi)=2i$ if and only if $x=\frac\pi2+\pi n$ for some $n\in\mathbb Z$ and $y=-(-1)^n\operatorname{arsinh}(2)$.