Finding range of $k$ in quadratic inequality 
Find all $k$ such that the inequality 
  $$2^{\frac{2\pi}{\arcsin x}}-2(k+2)2^{\frac{\pi}{\arcsin x}}+8k<0$$ 
  holds for at least one real $x$.

$\bf{My attempt}$ Let $2^{\frac{\pi}{\arcsin x}}=t>0.$ Then $t^2-2(k+2)t+8k<0$
$$t^2-2(k+2)t+(k+2)^2+8k-(k+2)^2<0$$
$$\bigg(t-(k+2)\bigg)^2<(k+2)^2-8k=(k-2)^2$$
$$-(k-2)<t-(k+2)<k-2$$
So $$4<t<2k$$
Could some help me how to solve it, thanks
 A: Hint. Note that $t(x)=2^{\frac{\pi}{\arcsin(x)}}$ is defined in $D:=[-1,0)\cup (0,1]$ and 
$$t(D)=(0,1/4]\cup[4,+\infty).$$
Moreover, as you already remarked.
$$t^2-2(k+2)t+8k=(t - 4) (t - 2 k)<0\Leftrightarrow\mbox{$t$ is strictly between $4$ and $2k$}$$
Now consider three cases:
i) if $k>2$ then $t(D)\cap (4,2k)\not=\emptyset$ and the required $x$ exists; 
ii) if $1/8\leq k\leq 2$ then $t(D)\cap (2k,4)\not=\emptyset$?
iii) if $k<1/8$ then $t(D)\cap (2k,4)\not=\emptyset$?
Can you take it from here?
A: Okay so we get $4<t<2k$ when we assume $k>2$. Let $k=2^{\gamma}$, then we have that
$$2^2<2^{\pi/\arcsin(x)}<2^{1+\gamma}$$
thus
$$2<\pi/\arcsin(x)<1+\gamma$$
We know that $\text{Range}(\pi/\arcsin(x))=\{y:|y|\ge 2\}$. Thus there exists such an $x$ if and only if $(2, 1+\gamma)\cap\{y:|y|\ge 2\}\neq\emptyset$, which is to say $\gamma > 1$. Thus if $k>2$, we find all $k>2$ work. Now assume that $k\le 2$, then
$$-(2-k)<t-(k+2)<2-k$$
so $2k<t<4$. Since $t>0$, this works for all non-positive $k$. Otherwise $k>0$ and write $k=2^{\gamma}$, then
$$1+\gamma<\pi/\arcsin(x)<2$$
And similarily we find this works when $\gamma<-3$. So when $k<2$ the valid options are $k\le 0$ and $k< 2^{-3}$. So in conclusion, $k\in (-\infty, 2^{-3})\cup (2, \infty)$.
