Domain of functions involving arcsine or arccosine

Question

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...

Answer 1

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, $$\implies \frac{x+1}{4} \ne \cos(-2)$$ $$\implies x \ne 4\cos(-2) -1$$