The domain of the function $$f(x) = \sin^{-1}[2x^2 - 3] + \log_2\left(\log_{\frac{1}{2}}(x^2 - 5x + 5)\right)$$, where $$[t]$$ is the greatest integer function, is

Given,

$$f(x)=\sin^{-1}[2x^2-3]+\log_2\left(\log_{\frac12}(x^2-5x+5)\right)$$

For the function to exist, both parts must be defined.

First consider

$$\sin^{-1}[2x^2-3]$$

Since $$\sin^{-1}t$$ is defined for $$-1\le t\le1,$$

we need $$-1\le[2x^2-3]\le1$$

As the GIF gives only integers,

$$[2x^2-3]\in\{-1,0,1\}$$

Hence,

$$-1\le2x^2-3<2$$

$$2\le2x^2<5$$

$$1\le x^2<\frac52$$

Therefore,

$$x\in\left(-\sqrt{\frac52},-1\right]\cup\left[1,\sqrt{\frac52}\right)$$

Now consider

$$\log_2\left(\log_{\frac12}(x^2-5x+5)\right)$$

For the outer logarithm,

$$\log_{\frac12}(x^2-5x+5)>0$$

Since the base

$$\frac12<1,$$

$$\log_{\frac12}t>0\iff0<t<1$$

Thus,

$$0<x^2-5x+5<1$$

Now solve separately.

From $$x^2-5x+5>0,$$

roots are $$x=\frac{5\pm\sqrt5}{2}$$

Hence,

$$x\in\left(-\infty,\frac{5-\sqrt5}{2}\right)\cup\left(\frac{5+\sqrt5}{2},\infty\right)$$

Also,

$$x^2-5x+5<1$$

$$x^2-5x+4<0$$

$$(x-1)(x-4)<0$$

$$1<x<4$$

Combining both,

$$x\in\left(1,\frac{5-\sqrt5}{2}\right)\cup\left(\frac{5+\sqrt5}{2},4\right)$$

Now intersect with

$$x\in\left(-\sqrt{\frac52},-1\right]\cup\left[1,\sqrt{\frac52}\right)$$

Since

$$\sqrt{\frac52}<\frac{5+\sqrt5}{2},$$

only the first interval survives.

Therefore, the domain is

$$\boxed{\left(1,\frac{5-\sqrt5}{2}\right)}$$.