Answer
$x=n\pi$ or $x=\frac{3\pi}{2}+2n\pi$, where $n$ is an integer.
Work Step by Step
$sin~x(sin~x+1)=0$
$sin~x=0$ or $sin~x=-1$
The period of $sin~x$ is $2\pi$. The solutions in the interval: $[0,2\pi)$ are:
$x=0$, $x=\pi$, $x=\frac{3\pi}{2}$
Now, add multiples of $2\pi$ to each of the solutions:
$x=0+2n\pi=2n\pi$
$x=\pi+2n\pi$
$x=\frac{3\pi}{2}+2n\pi$
We can rewrite the solutions $x=2n\pi$ and $x=\pi+2n\pi$ as:
$x=n\pi$
Finally:
$x=n\pi$ or $x=\frac{3\pi}{2}+2n\pi$, where $n$ is an integer.
