Solving $cos(x)=2$ Hi I was wondering if anyone can confirm my solution and also why I have a plus or minus in the answer? I calculated $z=2\pi n \pm 
i\cosh^{-1} (2)$  where $n \in \Bbb Z$ and if this is correct how would I go about plotting the solutions in an Argand diagram since the real compartment is $2\pi n$
 A: Your answer is correct, I think.
$$\cos (2\pi n \pm i \cosh^{-1}(2)) = \cos(2 \pi n) \cos (i \cosh^{-1}(2)) \pm \sin(2 \pi n) \sin (i \cosh^{-1}(2)) $$
Clearly, $\sin(2 \pi n)$ is always $0$ for $n \in$ integers, which allows for our $\pm$
Note that $\cosh(x) = \cos(ix)$
Therefore, 
$$\cos(2 \pi n) \cos (i \cosh^{-1}(2)) = 1 \cdot \cosh(\cosh^{-1}2) = 2$$
On an argand diagram, I believe you would simply have points along the y axis at $iy=i\cosh^{-1}2$ at intervals of $2 \pi$
A: By definition we know thay$$cos(x)=\frac{e^{ix}+e^{-ix}}{2}$$so$$\frac{e^{ix}+e^{-ix}}{2}=2\\e^{ix}+e^{-ix}=4$$now let's multiply both sides by $e^{ix}$
$$e^{ix^2}+1=4e^{ix}$$set $e^{ix}=u$ and you will get $$u^2-4u+2=0$$you will get$$\begin{cases}u_1=2-\sqrt3\\u_2=2+\sqrt3\end{cases}$$so$$e^{ix}=2\pm\sqrt3\\ix=\ln(2\pm\sqrt3)+2ni\pi\\x=\frac{\ln(2\pm\sqrt3)+2ni\pi}{i}\\=\frac{\ln(2\pm\sqrt3)}{i}+2n\pi$$this is the way to calculate the exact value, as for if your ways is okay? the definition of $\text{arccosh}$ is $$\text{arccosh}(z)=\frac{\sqrt{z-1}}{\sqrt{1-z}}\arccos(z)$$ so $$i \text{arccosh}(2)=i\frac{\sqrt{2-1}}{\sqrt{1-2}}\arccos(2)\\=i\frac1 i\arccos(2)=\arccos(2)$$ and by definition $\cos(\arccos(z))=z$ so you are right
