<s>[INST] Proving $ 2\arcsin\left(\sinh(x)\right) + \arccos\left( 2 - \cosh(2x) \right) = 0 $ 
$$ 2\arcsin\big( \sinh(x) \big) + \arccos\big( 2 - \cosh(2x) \big) = 0 $$

If this is true, then for what values of $x$?  (Clearly, the $\arcsin(\cdot)$ and $\arccos(\cdot)$ are not valid for every real argument.)
 [/INST] Let 
$$
L(x) = 2\arcsin(\sinh(x)) \\
R(x) = -\arccos(2-2\cosh(x))
$$
I'll show that these functions are equal for $x \ge 0$ when $|\sinh(x)| \le 1$. 
Step 1: check that $L(0) = R(0)$. Yes. 
Step 2: Check that $L'(x) = R'(x)$ for all $x$. 
Start from the following: for $0 \le u \le 1$, we have
$$
\sqrt{4u^2 - 4u^4} = 2u \sqrt{1-u^2},
$$
because $u > 0$, so we don't need the absolute value sign on the right (although including it would probably deal with the case where $x < 0$ later, but I leave that to OP). 
Then
\begin{align}
\sqrt{1 - (1 - 4u^2 + 4u^4)} &= 2u \sqrt{1-u^2} \\
\sqrt{1 - (1 - 2u^2)^2} &= 2u \sqrt{1-u^2}. 
\end{align}
If we replace $u$ with $\sinh(x)$, we get
\begin{align}
\sqrt{1 - (1 - 2\sinh^2(x))^2} &= 2\sinh(x) \sqrt{1-\sinh^2(x)}\\
\sqrt{1 - (2 - \cosh^2 (x) - \sinh^2(x))^2} &= 2\sinh(x) \sqrt{1-\sinh^2(x)}\\
\cosh(x)\sqrt{1 - (2 - \cosh^2 (x) - \sinh^2(x))^2} &= 2\cosh(x)\sinh(x) \sqrt{1-\sinh^2(x)}\\
\cosh(x)\sqrt{1 - (2 - \cosh^2 (x) - \sinh^2(x))^2} &= \sinh(2x) \sqrt{1-\sinh^2(x)}\\
2\cosh(x)\sqrt{1 - (2 - \cosh^2 (x) - \sinh^2(x))^2} &= 2\sinh(2x) \sqrt{1-\sinh^2(x)}\\
\frac{2\cosh(x)}{\sqrt{1-\sinh^2(x)}} &= \frac{2\sinh(2x)}{\sqrt{1 - (2 - \cosh^2 (x) - \sinh^2(x))^2}}\\
\frac{2\cosh(x)}{\sqrt{1-\sinh^2(x)}} &= \frac{2\sinh(2x)}{\sqrt{1 - (2 - \cosh(2x))^2}}\\
\end{align}
The left hand side of this is just $L'(x)$; the right hand side is $R'(x)$. Because they're equal, and $L(0) = R(0)$, we have that $L(x) = R(x)$ within the specified domain (by the FTC, if you want to get fancy). 
In short: the OP's statement, with a sign-fix, is correct for positive $x$; a similar proof, without the sign-change, presumably demonstrates the original claim for negative $x$. The whole thing is an exercise in algebra and the FTC. 
</s>