For some angle $x$, the equation

$$
a \cos ^{2} x+b \cos x+c=0
$$

holds. Write down the quadratic equation that $\cos 2 x$ satisfies. Apply your result to the case $a=4, b=2$, $c=-1$.
I. solution: The unknown of the given equation is connected to the unknown of the sought equation by the following identity:

$$
\cos x=\sqrt{\frac{1+\cos 2 x}{2}}, \text { or } 2 \cos ^{2} x=1+\cos 2 x
$$

(where the sign of the square root must be determined separately based on which quadrant $x$ falls into). Therefore, the sought equation can be obtained by substituting (2) into (1) and removing the square root from the resulting equation. This procedure is the same as expressing the first power of $\cos x$ from (1) in terms of the coefficients and $\cos ^{2} x$, squaring it, and then substituting $\cos ^{2} x$ according to the second identity in (2); step by step,

$$
\begin{gathered}
-b \cos x=a \cos ^{2} x+c \\
b^{2} \cos ^{2} x=a^{2} \cos ^{4} x+2 a c \cos ^{2} x+c^{2} \\
\frac{b^{2}-2 a c}{2}(1+\cos 2 x)=\frac{a^{2}}{4}\left(1+2 \cos 2 x+\cos ^{2} 2 x\right)+c^{2}
\end{gathered}
$$

and the sought equation is:

$$
\frac{1}{4} a^{2} \cos ^{2} 2 x+\frac{1}{2}\left(a^{2}-b^{2}+2 a c\right) \cos 2 x+\frac{1}{4}\left(a^{2}+4 a c+4 c^{2}-2 b^{2}\right)=0
$$

With the given triplet $a, b, c$, (1) and (3) become:

$$
\begin{gathered}
4 \cos ^{2} x+2 \cos x-1=0 \\
4 \cos ^{2} 2 x+2 \cos 2 x-1=0
\end{gathered}
$$

thus $\cos x$ and $\cos 2 x$ are roots of the same equation. The surprising fact that the two equations have the common roots $(-1 \pm \sqrt{5}) / 4$ means that the values of $x$ in the interval $\left(-180^{\circ}, 180^{\circ}\right)$ are

$$
-144^{\circ},-72^{\circ}, 72^{\circ}, 144^{\circ}
$$

and the values of $2 x$ in the interval $\left(-360^{\circ}, 360^{\circ}\right)$, and then in the interval $\left(-180^{\circ}, 180^{\circ}\right)$ are

$$
-288^{\circ},-144^{\circ}, 144^{\circ}, 288^{\circ}, \text { i.e., } 72^{\circ},-144^{\circ}, 144^{\circ},-72^{\circ}
$$

thus the two quartets of roots differ only in the order of listing.

Jenő Székely (Pécs, Nagy Lajos Gymnasium III. grade)

II. solution: Let the roots of (1) be $(\cos x)_{1}=z_{1},(\cos x)_{2}=z_{2}$, and the roots of the sought

$$
A \cos ^{2} 2 x+B \cos 2 x+C=0
$$

equation be $(\cos 2 x)_{1}=w_{1},(\cos 2 x)_{2}=w_{2}$. Then, according to (2),

$$
z^{2}=\frac{1+w}{2}, \quad \text { from which } \quad w_{1}=2 z_{1}^{2}-1, \quad w_{2}=2 z_{2}^{2}-1
$$

With these, using the expressions for the coefficients in terms of the roots,

$$
\begin{aligned}
-\frac{B}{A} & =w_{1}+w_{2}=2\left(z_{1}^{2}+z_{2}^{2}\right)-2=\left[2\left(z_{1}+z_{2}\right)^{2}-2 z_{1} z_{2}\right]-2 \\
\frac{C}{A} & =w_{1} w_{2}=4 z_{1}^{2} \cdot z_{2}^{2}-2\left(z_{1}^{2}+z_{2}^{2}\right)+1=4 z_{1}^{2} \cdot z_{2}^{2}-2\left(z_{1}+z_{2}\right)^{2}+4 z_{1} z_{2}+1
\end{aligned}
$$

and since $z_{1}+z_{2}=-b / a$, and $z_{1} z_{2}=c / a$, we have

$$
\begin{gathered}
-\frac{B}{A}=\frac{2 b^{2}}{a^{2}}-\frac{4 c}{a}-2=\frac{2 b^{2}-4 a c-2 a^{2}}{a^{2}} \\
\frac{C}{A}=\frac{a^{2}+4 a c+4 c^{2}-2 b^{2}}{a^{2}}
\end{gathered}
$$

With these, choosing $A$ as $a^{2} / 4$, we again arrive at (3) from (4).

Peregrin Timár (Budapest, Eötvös J. Gymnasium IV. grade)

Remarks. 1. Several have examined under which value systems of $a, b, c$ (1) has (real) solutions, i.e., such that the roots $z_{1,2}=\left(-b \pm \sqrt{b^{2}-4 a c}\right) / 2 a$ fall between -1 and 1, including these boundaries, and thus there is an angle $x$ for which (1) holds. In this, we can assume that $a>0$. - First of all, the

$$
b^{2}-4 a c \geq 0
$$

inequality must hold, because only then are $z_{1}$ and $z_{2}$ real. This means that the lowest point of the graph of the function $y=a z^{2}+b z+c$, considered as a function of $\cos x=z$, lies below, or on, the $Z$-axis (the abscissa axis). Indeed, due to symmetry, the abscissa of the lowest point is the arithmetic mean of the roots (zeros) $z_{0}=\left(z_{1}+z_{2}\right) / 2=-b / 2 a$, and thus its ordinate is: $a(-b / 2 a)^{2}+b(-b / 2 a)+c=-\left(b^{2}-4 a c\right) / 4 a$, which is negative or 0 according to (7).

Both roots fall between -1 and 1 if and only if, on the one hand, $z_{0}$ falls between these limits:

$$
-1b^{2}-4 a c, \quad b^{2}-4 a c0, \quad\left(9^{\prime}\right) \quad \text { or } \quad a+b+c>0
$$

$\left(9^{\prime}\right)$ and $\left(9^{\prime \prime}\right)$ state that the value of the function $f(z)$ at $z=-1$, and at $z=1$, is positive, or in the graphical representation of the parabola, the points $(-1 ; a-b+c)$ and $(1 ; a+b+c)$ are both above the $X$-axis.

If (7), (8), ( $9^{\prime}$ ), and ( $9^{\prime \prime}$ ) already hold, i.e., $z_{1}$ and $z_{2}$ are real and fall between -1 and 1, then (5) implies that the same holds for $w_{1}$ and $w_{2}$. It is obvious that in this case $w_{1}$ and $w_{2}$ are real, and if for $i=1,2$

$$
-1(a-c)^{2}