<s>[INST] What is the domain of the function $f(x)=\sin^{-1}\left(\frac{8(3)^{x-2}}{1-3^{2(x-1)}}\right)$? 
What is the domain of the function
  $f(x)=\sin^{-1}\left(\frac{8(3)^{x-2}}{1-3^{2(x-1)}}\right)$?

I started using the fact  $-\frac{\pi}{2}\le f(x) \le \frac{\pi}{2}\implies-1 \le\frac{8(3)^{x-2}}{1-3^{2(x-1)}}\le1$.Now,on dissecting it into two  cases.
$CASE (1): -1 \le\frac{8(3)^{x-2}}{1-3^{2(x-1)}}$ 
$CASE (2): \frac{8(3)^{x-2}}{1-3^{2(x-1)}}\le1$ 
The calculations is inboth cases are bewidering,that's why i'm not showing it.
I need someone who can help me in solving this.
 [/INST] Examine that,
$$\lim_{x\to-\infty}\frac{8(3)^{x-2}}{1-3^{2(x-1)}}=0$$
$$\lim_{x\to\infty}\frac{8(3)^{x-2}}{1-3^{2(x-1)}}=0$$
Also the function is discontinuous at $x=1$.
Now check where it assumes the values $1$ and $-1$.
$$\frac{8(3)^{x-2}}{1-3^{2(x-1)}}=1$$
$$8(3)^{x-2}=1-3^{2(x-1)}$$
$$8(3)^{x-2}+3^{2(x-1)}=1$$
$$3^{x-1}(\frac{8}{3}+3^{x-1})=1$$
Let $3^{x-1}=t$
$$t(8+3t)=3$$
$$t=\frac{1}{3} \implies x=0 $$ 

$$\frac{8(3)^{x-2}}{1-3^{2(x-1)}}=-1$$
$$8(3)^{x-2}=-1+3^{2(x-1)}$$
$$8(3)^{x-2}-3^{2(x-1)}=-1$$
$$3^{x-1}(\frac{8}{3}-3^{x-1})=-1$$
Let $3^{x-1}=t$
$$t(8-3t)=-3$$
$$t=3 \implies x=2 $$ 

Therefore, Domain is, $$(-\infty,0]\cup[2,\infty)$$
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