9. Let $-\frac{\pi}{2} \leqslant x \leqslant \frac{\pi}{2}$, and the equation $\cos 2x - 4a \cos x - a + 2 = 0$ has two distinct solutions. Find the range of values for $a$.
Solve
$2 \cos ^{2} x-4 a \cos x-a+1=0$, let $f(t)=2 t^{2}-4 a t-a+1$,
since $-\frac{\pi}{2} \leqslant x \leqslant \frac{\pi}{2} \Rightarrow t=\cos x \in[0,1]$, then $f(t)$ has exactly one root in the interval $[0,1)$. When $f(0)=0$, $a=1 \Rightarrow f(t)=2 t^{2}-4 t$ satisfies the condition;
When $f(0) \neq 0$, $f(0) \cdot f(1)<0 \Rightarrow(-a+1)(2-5 a+1)<0 \Rightarrow \frac{3}{5}<a<1$; $\Delta=16 a^{2}+8(a-1)=0 \Rightarrow 2 a^{2}+a-1=0 \Rightarrow a=\frac{1}{2}$ also satisfies the condition.
In summary, the range of $a$ is $\left\{\frac{1}{2}\right\} \cup\left(\frac{3}{5}, 1\right]$.