China Mathematical Competition

$a+a^2\ge 2$ when $x=0$. So $a\le -2$ (because $a<0$). When $a\le -2$, we have $$\begin{array}{rl}{a}^2+a\cos x& \ge a^2+a\ge 2\ge {\cos }^{2}x+\cos x\\ & =1+\cos x-{\sin }^{2}x,\end{array}$$ that is, $${\sin }^{2}x+a\cos x+a^2\ge 1+\cos x.$$ Hence, the range of values for negative $a$ is $a\le -2$.