Answer
$\frac{5\pi}{4}+2k\pi$, $\frac{7\pi}{4}+2k\pi$
Work Step by Step
$\begin{align}
\sqrt{2}\sin\theta+1&=0\\
\sqrt{2}\sin\theta&=-1\\
\sin\theta&=-\frac{1}{\sqrt 2}\\
\sin\theta&=-\frac{\sqrt{2}}{2}
\end{align}$
Since $\sin \theta$ is negative in the third and fourth quadrants, the only values of $\theta$ between $0$ and $2\pi$ that satisfy the equation are $\frac{5\pi}{4}$ and $\frac{7\pi}{4}$.
Since $\sin \theta$ is a periodic function with period $2\pi$, we can add or subtract any multiple of $2\pi$ from $\frac{5\pi}{4}$ or $\frac{7\pi}{4}$ and the equation would still be satisfied.
So the answers are $\frac{5\pi}{4}+2k\pi$ and $\frac{7\pi}{4}+2k\pi$ for any integer $k$.
