7. (2002 National High School Competition Question) The range of negative values of $a$ for which the inequality $\sin ^{2} x+a \cdot \cos x+a^{2}>1+\cos x$ holds for all $x \in \mathbf{R}$ is $\qquad$ .
7. Fill in $a \leqslant-2$. Reason: The original inequality can be transformed into $\left(\cos x-\frac{a-1}{2}\right)^{2} \leqslant a^{2}+\frac{(a-1)^{2}}{4}$. Since $-1 \leqslant \cos x \leqslant 1, a<0$, $\frac{a-1}{2}<0$, thus, when $\cos x=1$, the function $y=\left(\cos x-\frac{a-1}{2}\right)^{2}$ has a maximum value $\left(1-\frac{a-1}{2}\right)^{2}$. Therefore, we have $\left(1-\frac{a-1}{2}\right)^{2} \leqslant a^{2}+\frac{(a-1)^{2}}{4}$, solving this gives $a \geqslant 1$ or $a \leqslant-2$, but since $a<0$, hence $a \leqslant-2$ is the solution.