<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] Hint:
Let $3^{x-2}=a\implies a>0$ for real $x$
$$-1\le\dfrac{8a}{1-9a^2}\le1$$
$$\dfrac{8a}{1-9a^2}\le1\iff0\le\dfrac{9a^2+8a-1}{9a^2-1}=\dfrac{(9a-1)(a+1)}{(3a-1)(3a+1)}$$
As $a.0,$ we need $\dfrac{9a-1}{3a-1}\ge0$
$\implies$ either $9a-1=0\iff a=?$ or $ a>$max$\left(\dfrac19,\dfrac13\right)$ or $a<$min$\left(\dfrac19,\dfrac13\right)$
$\implies x<-2$ or $x\ge-1$
Similarly for $$-1\le\dfrac{8a}{1-9a^2}\iff\dfrac{8a+1-9a^2}{1-9a^2}\ge0$$
$$\iff\dfrac{(9a+1)(a-1)}{(3a+1)(3a-1)}\le0$$
As $a.0,$ we need $\dfrac{a-1}{3a-1}\ge0$
Can you take it from here?
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