\section*{Problem 4 - 031114}

Determine all real \(x\) for which \(\sin ^{2} x+\sin ^{2} 2 x>\sin ^{2} 3 x\) holds!
}

According to the addition theorems, we have

\[
\begin{aligned}
\sin ^{2} 3 x & =(\sin x \cos 2 x+\cos x \sin 2 x)^{2} \\
& =\sin ^{2} x \cos ^{2} 2 x+\cos ^{2} x \sin ^{2} 2 x+2 \sin x \cos x \sin 2 x \cos 2 x \\
& =\sin ^{2} x\left(1-\sin ^{2} 2 x\right)+\left(1-\sin ^{2} x\right) \sin ^{2} 2 x+\sin ^{2} 2 x \cos 2 x \\
& =\sin ^{2} x+\sin ^{2} 2 x-2 \sin ^{2} x \sin ^{2} 2 x+\sin ^{2} 2 x\left(1-2 \sin ^{2} x\right)
\end{aligned}
\]

Thus, the real \(x\) that satisfy the inequality \(\sin ^{2} x+\sin ^{2} 2 x>\sin ^{2} 3 x\) also satisfy the inequality \(2 \sin ^{2} x \sin ^{2} 2 x>\sin ^{2} 2 x\left(1-2 \sin ^{2} x\right)\). If \(x\) is a multiple of \(\frac{\pi}{2}\), this inequality is never satisfied. Otherwise, after dividing by \(\sin ^{2} x\),

\[
2 \sin ^{2} x>1-2 \sin ^{2} x \Leftrightarrow|\sin x|>\frac{1}{2}
\]

Since \(x\) is not a multiple of \(\frac{\pi}{2}\), all

\[
x \in\left(k \pi-\frac{\pi}{2}, k \pi-\frac{\pi}{6}\right) \cup\left(k \pi+\frac{\pi}{6}, k \pi+\frac{\pi}{2}\right), k \in \mathbb{Z}
\]

satisfy the inequality.