Let real numbers $a, b, c, x$ satisfy
$$
a \cos ^{2} x+b \cos x+c=0 .
$$

Try to give a quadratic equation that $\cos 2 x$ satisfies in terms of $a, b, c$. Compare these two equations in the case of $a=4, b=2, c=-1$.
Given that $\cos ^{2} x=\frac{\cos 2 x+1}{2}$, transform equation (1) into $a \cos ^{2} x+c=-b \cos x$. Squaring both sides is natural: $a^{2} \cos ^{4} x+2 a c \cos ^{2} x+c^{2}=b^{2} \cos ^{2} x$, which simplifies to $a^{2} \cos ^{4} x+\left(2 a c-b^{2}\right) \cos ^{2} x+c^{2}=0$. Substituting $a^{2}\left(\frac{\cos 2 x+1}{2}\right)^{2}+\left(2 a c-b^{2}\right) \frac{\cos 2 x+1}{2}+c^{2}=0$, we get
$$
a^{2} \cos ^{2} 2 x+2\left(a^{2}+2 a c-b^{2}\right) \cos 2 x+(a+2 c)^{2}-2 b^{2}=0 .
$$

When $a=4, b=2, c=-1$, equations (1) and (2) become
$$
\begin{array}{l}
4 \cos ^{2} x+2 \cos x-1=0, \\
4 \cos ^{2} 2 x+2 \cos 2 x-1=0 .
\end{array}
$$

Their coefficients are the same. In this case, the solutions to equation (3) are $\cos x=\frac{-1 \pm \sqrt{5}}{4}$, which gives $x_{1}=k \cdot 360^{\circ} \pm 72^{\circ}$, $x_{2}=k \cdot 360^{\circ} \pm 144^{\circ}$, where $k$ is an integer. For
$$
2 x_{1}=2 k \cdot 360^{\circ} \pm 144^{\circ}, 2 x_{2}=2 k \cdot 360^{\circ} \pm 288^{\circ}
$$

we have $\cos 2 x_{1}=\frac{-\sqrt{5}-1}{4}$, $\cos 2 x_{2}=\frac{\sqrt{5}-1}{4}$. They clearly satisfy equation (4).