Which is the quadratic equation whose roots are the squares of the roots of

$$
\sin 2 \alpha \cdot x^{2}-2(\sin \alpha+\cos \alpha) x+2=0
$$

Solve the thus obtained equation.
The roots of the given equation are:

$$
\begin{aligned}
& x_{1}=\frac{2(\sin \alpha+\cos \alpha)+2(\sin \alpha-\cos \alpha)}{4 \sin \alpha \cos \alpha}=\frac{1}{\sin \alpha} \\
& x_{2}=\frac{2(\sin \alpha+\cos \alpha)-2(\sin \alpha-\cos \alpha)}{4 \sin \alpha \cos \alpha}=\frac{1}{\sin \alpha}
\end{aligned}
$$

Thus, the roots of the sought quadratic equation are:

$$
z_{1}=\frac{1}{\cos ^{2} \alpha}, z_{2}=\frac{1}{\sin ^{2} \alpha}
$$

So the equation is

$$
\sin ^{2} \alpha \cos ^{2} \alpha \cdot z^{2}-z+1=0
$$

or

$$
\sin ^{2} 2 \alpha \cdot z^{2}-4 z+4=0
$$

from which

$$
z=\frac{2(1 \pm \cos 2 \alpha)}{\sin ^{2} 2 \alpha}
$$

and thus

$$
z_{1}=\frac{1}{\sin ^{2} \alpha}, z_{2}=\frac{1}{\cos ^{2} \alpha}
$$

(Dénes Aladár, Györ.)

The problem was also solved by: Barna D., Friedmann B., Goldziher K., and Kármán T., Manheim E., Orlowszky F., Posgay B., Schiffer H., Szabó K., Weisz J.