\section*{Problem 4 - 021214}

Determine all solutions of the equation

\(\cos ^{2} x \cdot \cos ^{2} 2 x \cdot \cos ^{2} 3 x+\sin ^{2} x \cdot \sin ^{2} 2 x \cdot \sin ^{2} 3 x=\cos ^{2} x \cdot \cos ^{2} 2 x+\cos ^{2} x \cdot \cos ^{2} 3 x+\cos ^{2} 3 x \cdot \cos ^{2} 2 x\) for \(0^{\circ} \leq x \leq 360^{\circ} \).
}

The given equation is first transformed using \(\sin ^{2}(x)=1-\cos ^{2}(x)\), where for brevity we write \(c_{1} \equiv \cos (x), c_{2} \equiv \cos (2 x), c_{3} \equiv \cos (3 x)\):

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

Using the addition theorems, we obtain: \(c_{2}^{2}=2 c_{1}^{2}-1\) and \(c_{3}=4 c_{1}^{3}-3 c_{1}\). Substituting these into (1) yields:

\[
\begin{aligned}
c_{1}^{2}+\left(2 c_{1}^{2}-1\right)^{2}+\left(4 c_{1}^{3}-3 c_{1}\right)^{2} & =1 \\
16 c_{1}^{6}-20 c_{1}^{4}+6 c_{1}^{2}=2 c_{1}^{2}\left(8 c_{1}^{4}-10 c_{1}^{2}+3\right) & =0
\end{aligned}
\]

From \(c_{1}^{2}=0\) we obtain the first two solutions: \(x_{1}=90^{\circ}, x_{2}=270^{\circ}\).

For further solutions, the expression in the parentheses in (2) must vanish, leading to a biquadratic equation with the solutions \(c_{1}^{2}=\frac{5}{8} \pm \frac{1}{8}\).

From \(c_{1}^{2}=\frac{3}{4}\) we obtain the four solutions: \(x_{3}=30^{\circ}, x_{4}=150^{\circ}, x_{5}=210^{\circ}\) and \(x_{6}=330^{\circ}\), and from \(c_{1}^{2}=\frac{1}{2}\) four more solutions: \(x_{7}=45^{\circ}, x_{8}=135^{\circ}, x_{9}=225^{\circ}, x_{10}=315^{\circ}\).