Let's solve the trigonometric equation ${\sin }^2(x)+\sin (x)=0$:

We notice that both terms have a common factor of sin$x$, so we can factor out sin$x$ from the equation:

$\sin (x)(\sin (x)+1)=0$

Now, we have two possibilities:

1) $\sin (x)=0$

This occurs when x is a multiple of π:

$x=n\pi$ where n is an integer.

2) $\sin (x)+1=0$

This implies that $\sin (x)=-1$, which occurs when x is in the form:

$x=(2n+1)\frac{\pi }{2}$ where n is an integer.

Therefore, the solutions to the equation ${\sin }^2(x)+\sin (x)=0$ are:

$x=n\pi$ and $x=(2n+1)\frac{\pi }{2}$ where n is an integer.