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

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

closed as off-topic by Saad, jvdhooft, ccorn, pisco, Jean-Claude ArbautJun 8 '18 at 10:55

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Saad, jvdhooft, ccorn, pisco, Jean-Claude Arbaut
If this question can be reworded to fit the rules in the help center, please edit the question.

• Do you mean $x^2-2x<\sin^{-1}(\sin 2)$? – Lord Shark the Unknown Jun 8 '18 at 6:10
• @LordSharktheUnknown edit my question please I am new in this platform and don't know how to write expression perfectly – Mohit Jun 8 '18 at 6:12

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