Answer
$135^{\circ}; 315^{\circ}$
Work Step by Step
Recall: $\sin^{2}\theta+\cos^{2}\theta=1$
Knowing that $\cos\theta=-\sin\theta$, we get
$ \cos^{2}\theta+\cos^{2}\theta=1$
$\implies 2\cos^{2}\theta=1$
$\implies \cos^{2}\theta=\frac{1}{2}$ or $\cos\theta=\pm\frac{1}{\sqrt 2}$
As sine and cosine function values have different signs in quadrant II and IV, we find the value of $\theta$ that lies in these quadrants.
$\cos45^{\circ}=\cos(-45^{\circ})=\frac{1}{\sqrt 2}$
The coterminal angle with $-45^{\circ}$ that is in between $0^{\circ}$ and $360^{\circ}$ is $(-45+360)^{\circ}=315^{\circ}$
$-\frac{1}{\sqrt 2}=-\cos 45^{\circ}=\cos(180^{\circ}-45^{\circ})=\cos 135^{\circ}$
$\theta=135^{\circ},315^{\circ}$
