14. Let $-\frac{\pi}{2} \leqslant x \leqslant \frac{\pi}{2}$, and the equation $\cos 2 x-4 a \cos x-a+2=0$

has two distinct solutions. Find the range of values for $a$.
14. From the given,
$2 \cos ^{2} x-1-4 a \cos x-a+2=0$,
which is $2 \cos ^{2} x-4 a \cos x-a+1=0$.
Let $t=\cos x$. From $-\frac{\pi}{2} \leqslant x \leqslant \frac{\pi}{2}$ we have
$0 \leqslant t \leqslant 1$.
Thus, the original equation having two distinct solutions is equivalent to the equation $2 t^{2}-4 a t-a+1=0$ having two distinct solutions in $0 \leqslant t \leqslant 1$.

This is equivalent to:
\[
\left\{\begin{array}{l}
f(0)>0, \\
f(1) \leqslant 0
\end{array} \Rightarrow\left\{\begin{array}{l}
-a+1>0, \\
-5 a+3 \leqslant 0
\end{array} \Rightarrow \frac{3}{5} \leqslant a<1\right.\right.
\];

or
\[
\left\{\begin{array}{l}
f(0)<0, \\
f(1) \geqslant 0
\end{array} \Rightarrow\left\{\begin{array}{l}
-a+1<0, \\
-5 a+3 \geqslant 0
\end{array}\right.\right.
\]
\[
\Rightarrow \text{ there is no } a \text{ satisfying the conditions };
\]

or
\[
\left\{\begin{array}{l}
0<a \leqslant 1, \\
2 a^{2}+a-1=0
\end{array} \Rightarrow\left\{\begin{array}{l}
0<a \leqslant 1, \\
(a+1)(2 a-1)=0
\end{array}\right.\right.
\]
\[
\Rightarrow\left\{\begin{array}{l}
0<a \leqslant 1, \\
a=-1 \text { or } a=\frac{1}{2}
\end{array} \Rightarrow a=\frac{1}{2}\right.
\].

Therefore, the range of $a$ is $\frac{3}{5} \leqslant a<1$ or $a=\frac{1}{2}$.