Domain of functions involving arcsine or arccosine I have two rather simple functions. I don't know how to find that the dominator is different than 0. Find the domain of functions:
Function 1:
$f(x) = \frac{4-x}{\arcsin\frac{x}{4}}$
Assumption 1:
$-1\leq \arcsin\frac{x}{4} \leq 1$
$-1\leq \frac{x}{4} \leq 1$
$-4\leq x \leq 4$
$x\in<4;4>$
Assumption 2:
$\arcsin\frac{x}{4} \neq 0$
Here, I have no idea how to proceed further with assumption 2.
Function 2:
$f(x) = \frac{\sqrt{2x-1}}{2+\arccos\frac{x+1}{4}}$
Assumption 1:
$\sqrt{2x-1} \geq 0$
$2x \geq 1$
$x \geq \frac{1}{2}$
Assumption 2:
$ -1 \leq \arccos \frac{x+1}{4} \leq 1 $
$ -1 \leq \frac{x+1}{4} \leq 1 $
$ -4 \leq x + 1 \leq 4 $
$ -5 \leq x \leq 3 $
Assumption 3:
$ 2 + \arccos\frac{x+1}{4} \neq 0 $
Here once again, no idea how to proceed further. 
Would anyone be able to help me understand these examples, how to find that the denominator is not equal to $0$, when there's arccos or arcsin in the denominator? Thanks! I do hope my tags are okay...

For the first function, the fraction is defined when $\arcsin \frac{x}{4} \ne 0 \implies \frac{x}{4} \ne \sin 0=0 \implies x\ne 0$
So the domain of the first function is $[-4,4]-\{0\}$
For the second function $$2+\arccos \frac{x+1}{4} \ne 0$$
$$\implies \arccos \frac{x+1}{4} \ne -2$$
Apply cos both sides,
$$\implies \frac{x+1}{4} \ne \cos(-2)$$
$$\implies x \ne 4\cos(-2) -1$$
