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