#### Answer

$y=\displaystyle \frac{3\pi}{4}$

#### Work Step by Step

$y=\sec^{-1}x =$arcsec$x$
Domain: $(-\infty, -1]\cup[1, \infty)$
Range: $[0,\displaystyle \frac{\pi}{2})\cup(\frac{\pi}{2},\pi]$
Quadrants (unit circle): I and II
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$y$ is the number from $[0,\displaystyle \frac{\pi}{2})\cup(\frac{\pi}{2},\pi]$
such that $\sec y=-\sqrt{2}$
$(\displaystyle \cos y=\frac{1}{-\sqrt{2}}=-\frac{\sqrt{2}}{2})$
In quadrant I, $\displaystyle \cos\frac{\pi}{4}=\frac{\sqrt{2}}{2}$, and
in quadrant II, $\displaystyle \cos\frac{3\pi}{4}=-\frac{\sqrt{2}}{2}$,
that is,
$\displaystyle \sec\frac{3\pi}{4}=-\sqrt{2}, $
and
$\displaystyle \frac{3\pi}{4}\in[0,\frac{\pi}{2})\cup(\frac{\pi}{2},\pi]$.
So,
$y=\displaystyle \frac{3\pi}{4}$