The inequality ${\sin }^{-1}\left(\sin 5\right)>x^2-4x$ holds if

${\sin }^{-1}\left(\sin 5\right)>x^2-4x\Rightarrow {\sin }^{-1}\left(\sin (5-2\pi \right)>x^2-4x\Rightarrow 5-2\pi >x^2-4x$

$\Rightarrow 9-2x>{(x-2)}^2\Rightarrow {(x-2)}^2<9-2\pi$

$\Rightarrow -\sqrt{9-2\pi }<x-2<\sqrt{9-2\pi }\Rightarrow 2-\sqrt{9-2\pi }<x<2+\sqrt{9-2\pi }$