Answer
$ 225^{\circ}$ and $315^{\circ}$.
Work Step by Step
$\csc\theta=-\sqrt 2$
$\sin\theta=\frac{1}{\csc\theta}=-\frac{1}{\sqrt 2}$
The value of $\sin \theta$ is negative, so $\theta$ may lie in either quadrant III or IV.
We know that $\sin 45^{\circ}=\frac{1}{\sqrt 2}$.
Recall: $-\sin x= \sin(180^{\circ}+x)$ and $-\sin x= \sin(360^{\circ}-x)$
$\implies -\sin 45^{\circ}=-\frac{1}{\sqrt 2}=\sin(180^{\circ}+45^{\circ})$
$=\sin 225^{\circ}$
$225^{\circ}$ which is in the third quadrant is one value of $\theta$.
Now $-\sin 45^{\circ}=-\frac{1}{\sqrt 2}= \sin (360^{\circ}-45^{\circ})$
$=\sin 315^{\circ}$
$315^{\circ}$ is in the fourth quadrant.
The values of $\theta$ are $ 225^{\circ}$ and $315^{\circ}$.
