<s>[INST] Find the domain of $\arcsin\left(\frac{3-x}{\sqrt{9-x^2}}\right)$ 
Find the domain of
  $$\arcsin\left(\dfrac{3-x}{\sqrt{9-x^2}}\right)$$

This question is given in my book. The answer is as follows:

For the function to be defined, $-1\le3-x\le1$ and $9-x^2>0$. Solve these inequalities and take their common region.
  $$2\le x\le4\tag1$$
  $$-3 < x < 3\tag2$$

The answer is $2\le x<3$. My question is why have they not taken $-1\le\dfrac{3-x}{\sqrt{9-x^2}}\le 1$, in which case the answer would have been different. Please help.
 [/INST] For $x\ne3,\dfrac{3-x}{\sqrt{9-x^2}}=\sqrt{\dfrac{3-x}{3+x}}$
Now we need $$-1\le\sqrt{\dfrac{3-x}{3+x}}\le1\iff0\le\dfrac{3-x}{3+x}\le1$$
If $\dfrac{3-x}{3+x}=0\implies x=3\ \ \ \ (1)$
$\dfrac{3-x}{3+x}>0\iff(3-x)(3+x)>0\iff(x-3)(x+3)<0\iff-3<x<3\ \ \ \ (2)$
$\dfrac{3-x}{3+x}\le1\iff0\ge\dfrac{3-x}{3+x}-1=\dfrac{3-x-(3+x)}{3+x}=-\dfrac{2x}{3+x}$
By $(2),x+3>0,$ so we need $-2x\le0\iff x\ge0\ \ \ \ (3)$
By $(1),(2),(3):0\le x<3$
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