#### Answer

$\theta = 135^{\circ}$
$\theta = 225^{\circ}$

#### Work Step by Step

$sec~\theta = -\sqrt{2}$
$\frac{r}{x} = -\sqrt{2}$
We can let $x = -1$ and $r = \sqrt{2}$.
Then: $y^2 = r^2-x^2$
$y = \pm \sqrt{r^2-x^2}$
$x = \pm \sqrt{(\sqrt{2})^2-(-1)^2}$
$y = 1$ or $y = -1$
If $x = -1$ and $y = 1$, then $\theta$ makes an angle of $45^{\circ}$ above the negative x-axis. Then $\theta = 180^{\circ}-45^{\circ} = 135^{\circ}$
If $x = -1$ and $y = -1$, then $\theta$ makes an angle of $45^{\circ}$ below the negative x-axis. Then $\theta = 180^{\circ}+45^{\circ} = 225^{\circ}$