Answer
$\theta=45^{\circ}or \,135^{\circ}$
Work Step by Step
We find:
$2\sin\theta-\sqrt 2=0$
$\sin\theta=\frac{\sqrt 2}{2}=\frac{1}{\sqrt 2}$
$\theta=\sin^{-1}(\frac{1}{\sqrt 2})$
For, $0^{\circ}\leq\,x\lt360^{\circ}$
So, $\theta=45^{\circ}or \,135^{\circ}$
