\section*{Problem 4 - 021234}

For all positive angles \(x\), give all solutions for

\[
\frac{1}{\sin ^{2} x}-\frac{1}{\cos ^{2} x} \geq \frac{8}{3}
\]
}

Given that \(\sin ^{2} x>0\) and \(\cos ^{2} x>0\) can be assumed due to the formulation of the inequality (equality is not allowed), the inequality can be transformed by multiplication to

\[
\cos ^{2} x-\sin ^{2} x \geq \frac{8}{3} \sin ^{2} x \cos ^{2} x
\]

which, using addition theorems, is equivalent to

\[
\cos 2 x \geq \frac{2}{3} \sin ^{2} 2 x=\frac{2}{3}\left(1-\cos ^{2} 2 x\right)
\]

Substituting \(z=\cos 2 x\) reduces the problem to the quadratic inequality \(2 z^{2}+3 z-2 \geq 0\). Its solution set is \(z \leq-2 \cup z \geq \frac{1}{2}\), but only for \(z \in\left[\frac{1}{2}, 1\right]\) is back-substitution possible. This yields that

\[
x \in\left(0, \frac{\pi}{6}\right] \cup\left[\frac{5 \pi}{6}, \pi\right)
\]

must hold; the given inequality is only satisfied for these \(x\) (and their repetition with the period \(\pi\)).