4.90 Given a quadratic equation in terms of $\cos x$
$a \cos ^{2} x+b \cos x+c=0$, where $a, b, c$ are known real numbers. Find a quadratic equation whose roots are $\cos 2 x$. In the case of $a=4, b=2, c=-1$, compare the given equation with the newly derived equation.
[Solution] Transform the given equation into
$$a \cos ^{2} x+c=-b \cos x .$$

Square both sides to get
$$\begin{array}{l}
a^{2} \cos ^{4} x+2 a \cos ^{2} x \cdot c+c^{2}=b^{2} \cos ^{2} x \\
a^{2} \cos ^{4} x+\left(2 a c-b^{2}\right) \cos ^{2} x+c^{2}=0
\end{array}$$

Substitute $\cos ^{2} x=\frac{1+\cos 2 x}{2}$ into equation (1) to get
$$a^{2}\left(\frac{1+\cos 2 x}{2}\right)^{2}+\left(2 a c-b^{2}\right) \cdot \frac{1+\cos 2 x}{2}+c^{2}=0$$

Simplify to get
$$a^{2} \cos ^{2} 2 x+2\left(a^{2}+2 a c-b^{2}\right) \cos 2 x+(a+2 c)^{2}-2 b^{2}=0 .$$

This is a quadratic equation in terms of $\cos 2 x$.
In the case of $a=4, b=2, c=-1$, the given equation is
$$4 \cos ^{2} x+2 \cos x-1=0$$

The new equation is
$$4 \cos ^{2} 2 x+2 \cos 2 x-1=0$$

Thus, these two equations are quadratic equations with identical coefficients, where the first one has $\cos x$ as the unknown, and the second one has $\cos 2 x$ as the unknown.