\section*{Problem 4 - 041234}

For which real numbers \(x\) is the equation \(\tan ^{2} x+\cot ^{2} x=6\) satisfied?
}

Due to \(\tan (x+\pi)=-\cot x\) and \(\cot (x+\pi)=-\tan x\), we have

\[
\tan ^{2}\left(x+\frac{\pi}{2}\right)+\cot ^{2}\left(x+\frac{\pi}{2}\right)=\tan ^{2} x+\cot ^{2} x
\]

and because \(\tan \left(\frac{\pi}{4} \pm x\right)=\cot \left(\frac{\pi}{4} \mp x\right)\), we have

\[
\tan ^{2}\left(x+\frac{\pi}{4}\right)+\cot ^{2}\left(x+\frac{\pi}{4}\right)=\tan ^{2}\left(\frac{\pi}{4}-x\right)+\cot ^{2}\left(\frac{\pi}{4}-x\right)
\]

which means \(\tan ^{2} x+\cot ^{2} x\) is \(\frac{\pi}{2}\)-periodic and symmetric about \(x=\frac{\pi}{4}\). Therefore, we only need to find solutions of \(\tan ^{2} x+\) \(\cot ^{2} x=6\) in the interval \(I=\left(0, \frac{\pi}{4}\right)\) (0 excluded due to the undefined point). The given equation is

\[
\begin{array}{r}
\frac{\sin ^{2} x}{\cos ^{2} x}+\frac{\cos ^{2} x}{\sin ^{2} x}=6 \\
\Leftrightarrow \sin ^{4} x+\cos ^{4} x=6 \sin ^{2} x \cos ^{2} x \\
\Leftrightarrow\left(1-\cos ^{2} x\right)^{2}+\cos ^{4} x=6\left(1-\cos ^{2} x\right) \cos ^{2} x \\
\Leftrightarrow 2 \cos ^{4} x-2 \cos ^{2} x+1=6 \cos ^{2} x-6 \cos ^{4} x \\
\Leftrightarrow \cos ^{4} x-\cos ^{2} x+\frac{1}{8}=0 \\
\Leftrightarrow \cos ^{2} x=\frac{2 \pm \sqrt{2}}{4} \\
\Leftrightarrow \cos x=\frac{\sqrt{2 \pm \sqrt{2}}}{2}
\end{array}
\]

The last transformation is valid because \(\cos\) is positive on \(I\).

\(\cos x=\frac{1}{2} \sqrt{2 \pm \sqrt{2}}\) has no solution in \(I\), because \(\cos\) is strictly decreasing on \(I\) and \(\cos \frac{\pi}{4}=\frac{1}{2} \sqrt{2}>\) \(\frac{1}{2} \sqrt{2 \pm \sqrt{2}}\) holds. Therefore, the original equation has only one solution in \(I\), namely

\[
x=\arccos \frac{\sqrt{2+\sqrt{2}}}{2}=\frac{\pi}{8}
\]

The last equality holds because

\[
\cos \frac{\pi}{8}=\cos \left(\frac{1}{2} \cdot \frac{\pi}{4}\right)=\sqrt{\frac{1}{2}\left(1+\cos \frac{\pi}{4}\right)}=\sqrt{\frac{1}{2} \cdot \frac{2+\sqrt{2}}{2}}
\]

Due to the symmetry and periodicity mentioned at the beginning, the set of all real solutions is

\[
\left\{\frac{\pi}{8}, \frac{3 \pi}{8}\right\}+\frac{\pi}{2} \mathbb{Z}=\frac{\pi}{8}+\frac{\pi}{4} \mathbb{Z}=\left\{ \pm \frac{\pi}{8}, \pm \frac{3 \pi}{8}, \pm \frac{5 \pi}{8}, \ldots\right\}
\]