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.)
 A: 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. 
A: Recall a couple double-angle formulas valid for all $x$:
$$ \cos(2x)=1-2\sin^2(x), \\ \cosh(2x)=1+2\sinh^2(x). \tag{1}$$
Apply $u\mapsto 2-u$ to the latter so we may write
$$ 1-2\sinh^2(x)=2-\cosh(2x). \tag{2}$$
Use the substitution $\sinh(x)=\sin(\theta)$, where $\theta$ is in $[-\frac{\pi}{2},\frac{\pi}{2}]$, the range of $\arcsin$. This way the left side of $(2)$ is just $\cos(2\theta)$, so we may write this as
$$ \cos(2\theta)=2-\cosh(2x). \tag{3}$$
Split into two cases: either $2\theta$ is in $[0,\pi]$, the range of $\arccos$, or it's in $[-\pi,0]$.
In the first case, we can apply $\arccos$ to both sides of $(3)$ to obtain
$$ 2\theta=\arccos(2-\cosh(2x)) \tag{4a}$$
In the second case, we have $2\theta=-\arccos(\cos(2\theta))$, so from $(3)$ we get
$$ 2\theta=-\arccos(2-\cosh(2x)). \tag{4b}$$
The conditions on $\theta$ can be turned into conditions on $x$:
$$ \begin{array}{rrrrr} 0\le\theta\le\frac{\pi}{2} & \iff & 0\le \sin\theta\le 1 & \iff & 0\le x\le\sinh^{-1}(1) \\ -\frac{\pi}{2}\le\theta\le0 & \iff & -1\le\sin\theta\le0 & \iff & -\sinh^{-1}(1)\le x\le0 \end{array} \tag{5}$$
Finally, replace $\theta$ with $\arcsin(\sinh(x))$. In conclusion,
$$ \arcsin(\sinh(x))=\mathrm{sgn}(x)\arccos(2-\cosh(2x)) $$ for all $-\sinh^{-1}(1)\le x\le \sinh^{-1}(1).~$ (Note $\sinh^{-1}(1)=\ln(1+\sqrt{2})\approx 0.88$.)
A: Looks wrong to Desmos:

So...I guess that it's probably almost never true (except perhaps for $x = 0$). 
Details:
Let's look at $x = \frac{1}{2}$.
Now here's a matlab program to test that out:
function silly()
x = 0.5
a = sinh(x)
b = asin(a)
c = 2 * b

'part 2'
d = cosh(1)
f = 2 - d
g = acos(f)

'sum'
c + g

and the results of executing it:
>> silly
x =
    0.5000
a =
    0.5211
b =
    0.5481
c =
    1.0963
ans =
   'part 2'
d =
    1.5431
f =
    0.4569
g =
    1.0963
ans =
    'sum'
ans =
    2.1925

If $2.1925$ looks like zero to within roundoff error, then you are correct; if not, then Desmos and I are correct. It's telling that the two parts look identical, so that their difference at $x = 0.5$ would be zero, while their sum is nonzero. 
(To be fair, I should say that I was half-correct --- it's almost never true for positive $x$; it appears to be identically true for negative $x$.)
A: If we use the Taylor series, $\arcsin(\sinh(x))\approx x+\frac {x^3}3+\frac {x^5}6$ and $\arccos(2-2-\cosh(2x))\approx |2x+\frac {2x^3}3+\frac {x^5}3|$ so for $x \gt 0$ the dominant term is $4x$.  For $x \lt 0$ the first through fifth order terms cancel.
A: I suppose a typo; changing the $+$ to $-$ makes
$$2\arcsin\big( \sinh(x) \big) \color{red}{-} \arccos\big( 2 - \cosh(2x) \big) = 0$$  true for all $x \geq 0$.
Composing Taylor series built at $x=0$ and assuming $x >0$, we have
$$2\arcsin\big( \sinh(x) \big)=2 x+\frac{2 x^3}{3}+\frac{x^5}{3}+\frac{79 x^7}{315}+\frac{493
   x^9}{2268}+O\left(x^{11}\right)$$
$$\arccos\big( 2 - \cosh(2x) \big)=2 x+\frac{2 x^3}{3}+\frac{x^5}{3}+\frac{79 x^7}{315}+\frac{493
   x^9}{2268}+O\left(x^{11}\right)$$
