#### Answer

$\sin^{-1}2$ does not exist.

#### Work Step by Step

Solve for radians, then convert to degrees.
$y=\sin^{-1}x$
Domain: $[-1, 1] $
Range: $[-\displaystyle \frac{\pi}{2}, \displaystyle \frac{\pi}{2}]$
Quadrants (unit circle): I and IV
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$y$ is the number from $[-\displaystyle \frac{\pi}{2}, \displaystyle \frac{\pi}{2}]$
such that $\sin y=2.$
But,
such a y does not exist,
as sine can not be greater than 1.
More directly,
$2$ is not in the domain of $\sin^{-1}x$, so
$\sin^{-1}2$ does not exist.