Solve for (x) $x^2 - 2x <\sin^{-1}(\sin 2)$?

Question

I had tried but not able to get the solution help please I am getting an inequality

$x$^2-2$x$+2-π <0

And unable to factories

Answer 1

$x^{2}-2x$ < $sin^{-1}(sin(2))$ => $x^{2}-2x$ < $\pi-2$ =>$\frac{(x-1)^{2}}{\pi-1}$<1 then from the inequality we can deduce the solution as follows $1-\sqrt{\pi-1}<x<1+\sqrt{\pi-1}$