Find all solutions of the equation in the interval $[0,2 \pi)$. \[ \sin x=-\cos ^{2} x-1 \] Write your answer in radians in terms of $\pi$. If there is more than one solution, separate them with commas. \[ x= \]

Step 1 ：Since \(\sin x = \sqrt{1 - \cos^2 x}\), we get \(-\sqrt{1 - \cos^2 x} = -\cos^2 x - 1\).

Step 2 ：Squaring both sides, we get \(1 - \cos^2 x = \cos^4 x + 2\cos^2 x + 1\).

Step 3 ：Rearranging, we get \(\cos^4 x + 2\cos^2 x - \cos^2 x + 2 = 0\), which simplifies to \(\cos^4 x + \cos^2 x + 2 = 0\).

Step 4 ：This equation has no real solutions for \(\cos x\), so there are no solutions for \(x\) in the given interval.

Step 5 ：Thus, the final answer is \(\boxed{\text{No Solution}}\).