If $ sin^{-1}(sin 5) > x^2 - 4x , $ then 

$x ∈ ( 2 - \sqrt{9-2\pi}, 2+\sqrt{9-2\pi} )$

The correct answer is Option 2: $x ∈ ( 2 - \sqrt{9-2\pi}, 2+\sqrt{9-2\pi} )$

We have, 

$\sin^{-1}(\sin 5) > x^2 - 4x  $

We know:

$\sin^{-1}(\sin θ)=θ$ only if $θ∈\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$

But 5 radians is not in $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$. So we use:

$\sin^{-1}(\sin 5) =5-2\pi$ (since $5∈[\pi,2\pi]$ and $\sin 5=\sin(2\pi-5)$)

Now the inequality becomes:

$5-2\pi>x^2-4x$

Step 2: Bring all terms to one side:

$⇒ 5 - 2 \pi > x^2 - 4x $

$⇒ x^2 - 4x - 5 + 2 \pi < 0 ⇒2 - \sqrt{9-2 \pi} < x < 2 + \sqrt{9-2 \pi}$