The given relation can be rewritten as a quadratic equation with respect to $\cos x$ by replacing $\sin^2 x$ with $1-\cos^2 x$: \begin{equation} \cos^2 x -\cos x - (a+1)=0 \end{equation} For this equation to be solvable, its determinant must be non-negative: $$(-1)^2 + 4\cdot(a+1) \ge 0\implies a\ge-\frac{5}{4}$$ Meanwhile, because $$-1 \le\cos x=\frac{1\pm \sqrt{1+4\cdot(a+1)}}{2} \le 1\implies -\frac{5}{4}\le a \le 1$$ Therefore, the answer is $\boxed{-\frac{5}{4}\le a \le 1}$. Alternative Solution The relation can be rewritten as $$a = \cos^2 x -\cos x -1 = \Big(\cos x -\frac{1}{2}\Big)^2 -\frac{5}{4}$$ Therefore, the minimal value of $a$ is $0-\frac{5}{4}=-\frac{5}{4}$ and the maximum value equals $\Big(-1-\frac{1}{2}\Big)^2 -\frac{5}{4}=1$.