Answer
$\theta=\frac{3\pi}{2}+2k\pi$
Work Step by Step
We can first find all solutions in the interval $[0, 2\pi)$:
$\sin(\theta)+1=0$
$\sin(\theta)=-1$
$\theta=\frac{3\pi}{2}$
Because $\sin(\theta)$ has period $2\pi$, we can add any multiple of $2\pi$ to $\frac{3\pi}{2}$ and the equation would still hold:
$\theta=\frac{3\pi}{2}+2k\pi$ for any integer k
