# Solving $\cos(z) = \frac{5}{2}$

I'm given $$\cos(z) = \frac{5}{2}$$ and I'm trying to solve for $$z$$ but I keep going in circles. I know $$\cos z = 5/2 = 1/2(e^{iz}+e^{-iz})$$ so then $$e^{iz}+e^{-iz} = 5$$ but then I'm stuck

Taking $$t=e^{iz}$$ we get $$t+\frac{1}{t}=5 \implies t^2-5t+1=0 \implies t_{1,2} = \frac{5 \pm \sqrt{21}}{2}$$

Note

$$\cos z = \cosh(-i z +i 2\pi n)=\frac52$$

which leads to $$-i z +i 2\pi n=\pm \cosh^{-1}\frac52$$ and the solutions

$$z= 2\pi n\pm i \cosh^{-1}\frac52$$

$$z= \pm i \log\;(\dfrac{5 + \sqrt{21}}{2})$$ which is pure imaginary to which we add the real variable part $$2 \pi n$$ making up the complex angle.
Graphs of $$(\cos x, \frac52)$$ do not intersect. It is interesting to note however that the real part corresponds to the closest points between the non-intersecting cosine curve and straight line.