\section*{Problem 6 - 061236}

The number \(\sin 10^{\circ}\) satisfies an algebraic equation of the third degree with integer coefficients. Set up this (uniquely determined up to a common divisor of all coefficients) equation and determine its two other roots.
}

It is

\[
\begin{gathered}
\sin (2 x)=2 \sin (x) \cos (x) \quad, \quad \cos (2 x)=\cos ^{2}(x)-\sin ^{2}(x) \quad \text { and } \\
\sin (3 x)=\sin (2 x+x)=\sin (2 x) \cos (x)+\cos (2 x) \sin (x)= \\
=2 \sin (x) \cos ^{2}(x)+\cos ^{2}(x) \sin (x)-\sin ^{3}(x)=3\left(1-\sin ^{2}(x)\right) \sin (x)-\sin ^{3}(x)=3 \sin (x)-4 \sin ^{3}(x)
\end{gathered}
\]

With \(x=10^{\circ}, X_{1}=\sin (x)\) and \(\sin (3 x)=\sin \left(30^{\circ}\right)=\frac{1}{2}\) it follows that \(X_{1}=\sin \left(10^{\circ}\right)\) is a solution of the equation

\[
\frac{1}{2}=3 X-4 X^{3} \quad \text { or } \quad 8 X^{3}-6 X+1=0
\]

Since \(\sin \left(30^{\circ}\right)=\sin \left(390^{\circ}\right)=\sin \left(750^{\circ}\right)\), \(X_{2}=\sin \left(x_{2}\right)\) and \(X_{3}=\sin \left(x_{3}\right)\) with \(x_{2}=\frac{390^{\circ}}{3}=\) \(130^{\circ}\) and \(x_{3}=\frac{750^{\circ}}{3}=250^{\circ}\) also satisfy this equation.

Obviously, \(X_{1}=\sin \left(10^{\circ}\right), X_{2}=\sin \left(130^{\circ}\right)=\sin \left(50^{\circ}\right)\) and \(X_{3}=\sin \left(250^{\circ}\right)=\sin \left(-70^{\circ}\right)\) are pairwise distinct, since the sine function is strictly increasing in the interval \(\left[-90^{\circ}, 90^{\circ}\right]\). Therefore, \(X_{2}\) and \(X_{3}\) are the other solutions of the given equation.