2. 2.9 * The range of negative values of $a$ that make the inequality $\sin ^{2} x+a \cos x+a^{2} \geqslant 1+\cos x$ hold for all $x \in \mathbf{R}$ is $\qquad$ .
When $x=0$, $a+a^{2} \geqslant 2$, so $a \leqslant-2$ (since $a<0$). Also, when $a \leqslant-2$, we have
$$
a^{2}+a \cos x \geqslant a^{2}+a \geqslant 2 \geqslant \cos ^{2} x+\cos x=1+\cos x-\sin ^{2} x,
$$

which means $\sin ^{2} x+a \cos x+a^{2} \geqslant 1+\cos x$. Therefore, the range of $a$ is $a \leqslant-2$.
$$
f(\cos x)=\cos ^{2} x+(1-a) \cos x-a^{2} \leqslant 0(a<0) .
$$

Since $a-1<0$, the necessary and sufficient condition for $f(\cos x) \leqslant 0$ to hold for any $x \in \mathbf{R}$ is $f(1) \leqslant 0$, i.e., $1+(1-a)-a^{2} \leqslant 0$. Solving this, we get $a \leqslant-2$ (since $a<0$).