# 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

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• 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 $1-\sqrt{\pi-1}<x<1+\sqrt{\pi-1}$