A2 (1-3, Hungary) Given the quadratic equation in $\cos x$: $a \cos ^{2} x+b \cos x+c$ $=0$. Here, $a, b, c$ are known real numbers. Construct a quadratic equation whose roots are $\cos 2 x$. In the case where $a$ $=4, b=2, c=-1$, compare the given equation with the newly constructed equation.
Solve: Rewrite the original equation as
$$
a \cos ^{2} x+c=-b \cos x,
$$

square both sides and multiply by 4, to get
$$
2 b^{2}(1+\cos 2 x)=[a(1+\cos 2 x)+2 c]^{2},
$$

rearrange to obtain
$$
a^{2} \cos ^{2} 2 x+\left(2 a^{2}+4 a c-2 b^{2}\right) \cos 2 x+\left(a^{2}+4 a c-2 b^{2}+4 c^{2}\right)=0 \text {. }
$$

When $a=4, b=2, c=-1$, the above quadratic equation in $\cos 2 x$ becomes
$$
4 \cos ^{2} 2 x+2 \cos 2 x-1=0 \text {. }
$$

This has the same coefficients as the equation in $\cos x$.