Find all exact solutions on the interval $[0,2\pi$ ). Look for opportunities to use trigonometric identities. (Enter your answers as a comma-separated list. If an answer does not exist, enter DNE.)
$${\sin }^2(x)-{\cos }^2(x)-\sin (x)=0$$

Step 1 ：Given the equation ${\sin }^2(x)-{\cos }^2(x)-\sin (x)=0$.

Step 2 ：Use the Pythagorean identity ${\sin }^2(x)+{\cos }^2(x)=1$ to rewrite the equation as ${\sin }^2(x)-(1-{\sin }^2(x))-\sin (x)=0$.

Step 3 ：Simplify the equation to get $2{\sin }^2(x)-\sin (x)-1=0$.

Step 4 ：This is a quadratic equation in terms of $\sin (x)$, which can be solved using the quadratic formula to get the solutions $-1/2$ and $1$.

Step 5 ：Substitute these solutions back into the equation to find the corresponding values of $x$ in the interval $[0,2\pi )$. The solutions are $x=-0.523598775598299+2\pi$ and $x=1.57079632679490$.

Step 6 ：Convert these solutions from radians to degrees to get $x={330}^{\circ }$ and $x={90}^{\circ }$.

Step 7 ：Final Answer: The solutions to the equation ${\sin }^2(x)-{\cos }^2(x)-\sin (x)=0$ in the interval $[0,2\pi )$ are $x={330}^{\circ }$ and $x={90}^{\circ }$. So, the final answer is $\boxed{{\displaystyle {330}^{\circ },{90}^{\circ }}}$.