The following equation is given:

$$
(2 \cos \alpha-1) x^{2}-4 x+4 \cos \alpha+2=0
$$

where

$$
\alpha<90^{\circ}
$$

$1^{\circ}$. For which values of $\alpha$ are the roots of the equation real?

$2^{\circ}$. For these values of $\alpha$, what are the signs of the roots?

$3^{\circ}$. Transform the product of the roots to a form suitable for logarithmic use.
$1^{\circ}$. The roots of the equation are real if

$$
16-4(2 \cos \alpha-1) \cdot 2 \cdot(2 \cos \alpha+1)>0
$$

or

$$
3>4 \cos ^{2} \alpha
$$

from which

$$
\frac{\sqrt{3}}{2} \leqq \cos \alpha \leqq \frac{\sqrt{3}}{2}
$$

Since $\alpha > 0$, the inequality under (1) can be written as

$$
0 < \alpha \leqq 30^{\circ}
$$

$2^{\circ}$. The product of the roots is

$$
\frac{2(2 \cos \alpha+1)}{2 \cos \alpha-1}
$$

which is positive if the denominator is positive, i.e., if

$$
\cos \alpha>\frac{1}{2}
$$

or if

$$
\alpha < 60^{\circ}
$$

If $\alpha > 60^{\circ}$, the product of the roots is negative, so the roots have opposite signs. Therefore, both roots are positive if

$$
30^{\circ} \leqq \alpha < 60^{\circ}
$$

and the roots have opposite signs if

$$
60^{\circ} < \alpha < 90^{\circ}
$$

$3^{\circ}$.

$$
\begin{aligned}
\frac{4 \cos \alpha+2}{2 \cos \alpha-1} & =2 \cdot \frac{2 \sin \alpha \cos \alpha+\sin \alpha}{2 \sin \alpha \cos \alpha-\sin \alpha}=2 \frac{\sin 2 \alpha+\sin \alpha}{\sin 2 \alpha-\sin \alpha}= \\
& =\frac{4 \sin \frac{3}{2} \alpha \cos \frac{1}{2} \alpha}{2 \cos \frac{3}{2} \alpha \sin \frac{1}{2} \alpha}=2 \cdot \tan \frac{3}{2} \alpha \cdot \cot \frac{1}{2} \alpha
\end{aligned}
$$

(József Weisz.)

The problem was also solved by: B. Krausz, J. Krausz, Gy. Krisztián, Gy. Lukhaub, Ö. Spitzer.