4.90 Given the quadratic equation in $\cos x$: $a \cos ^{2} x+b \cos x+c=0$, where $a, b, c$ are known real numbers. Find a quadratic equation whose roots are $\cos 2 x$. Compare the original equation and the new equation in the case where $a=4, b=2, c=-1$.
[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 $\cos 2 x$.
For the case where $a=4, b=2, c=-1$,

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 is in terms of $\cos x$ and the second one is in terms of $\cos 2 x$.