Answer
$45^{\circ};225^{\circ}$
Work Step by Step
Recall: $\sin^{2}\theta+\cos^{2}\theta=1$
Given that $\sin\theta=\cos\theta$
$\implies \cos^{2}\theta+\cos^{2}\theta=1$
$\implies 2\cos^{2}\theta=1$
$\implies \cos^{2}\theta=\frac{1}{2}$ 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 quadrant I and III only.
$\cos^{-1}(\frac{1}{\sqrt 2})=45^{\circ}$
$-\frac{1}{\sqrt 2}=-\cos 45^{\circ}=\cos(180^{\circ}+45^{\circ})=\cos 225^{\circ}$
$\theta=45^{\circ},225^{\circ}$
