Let $a,b\in \mathbb{R}$. Consider the equation $$ 2a^2(1+\cos x) - (1+a^2 + b^2)^2 = 0\text{.} $$ The problem is to find all possible solutions $x \in \mathbb{C}$ (set of complex numbers) that satisfy the above equality for a given real numbers $a$ and $b$.
ATTEMPT: Solving for $x$, we arrived with $$ x = \arccos \Bigg[ \frac{(1+a^2 + b^2)^2}{2a^2}-1\Bigg]\text{.} $$ However, I got stuck because of the required domain of $\arccos(\cdot)$ which is $[-1,1]$ while the expression $\frac{(1+a^2 + b^2)^2}{2a^2}-1$ does not possibly be within this interval for arbitrary values of $a$ and $b$. Sometimes I am tempted to say that no general solution exists even in $\mathbb{C}$. Can anyone help me out?
EDIT: What I actually want is a simple detailed derivation for $x$ in terms of $a$ and $b$ which possibly be in complex form.
What you should know is $\arccos x$, if the domain is $[-1,1]$, it has real value, and if the domain isn't, then it has complex value.
Take a look: we calculate the value of $\cos i$, so using Euler formula:
$$\cos i=\dfrac{e^{ii}+e^{-ii}}{2}=\dfrac{e+e^{-1}}{2}$$
The answer is larger than $1$.
So if your problem is solved in $\mathbb{C}$, then there is no problem when the value inside $\arccos$ is not in the range $[-1,1]$.
Side note: You have solved your problem.