\section*{Problem 4 - 021134}

Determine all solutions of the equation \(\sin ^{3} x+\cos ^{3} x=1\).
}

First, one observes the following property of real numbers:

\[
\forall t \in \mathbb{R} \text { with } t1-\cos ^{2} x\) holds, except when \(\cos x=0\) or \(\cos x=1\), in which case equality holds. Similarly, \(\sin ^{2} x>\sin ^{3} x\) holds wherever \(\sin x\) is different from 0 and 1. Using the trigonometric Pythagoras in the form \(1-\cos ^{2} x=\sin ^{2} x\), it follows that \(1-\cos ^{3} x>\sin ^{3} x\), which in the form \(\sin ^{3} x+\cos ^{3} x<1\) is a direct contradiction to the equation whose solutions we are seeking.

Therefore, it can only have solutions where the inequality does not hold.

This is precisely the case where both \(\sin x\) and \(\cos x\) take one of the values 0 or 1, i.e., at \(x_{0}=0\) and \(x_{1}=\frac{\pi}{2}\). Indeed, these two values satisfy the equation, so the complete solution (taking periodicity into account) consists of all values

\[
x_{2 k}=2 k \pi \quad x_{2 k+1}=\frac{\pi}{2}+2 k \pi \quad(k \in \mathbb{N})
\]