\section*{Problem 6 - 081236}

All real numbers \(a\) are to be determined for which the equation

\[
\sin ^{6} x+\cos ^{6} x=a\left(\sin ^{4} x+\cos ^{4} x\right)
\]

has at least one real solution. Furthermore, all solutions for \(a=\frac{5}{6}\) are to be specified.
}

Using

\[
\sin ^{6} x+\cos ^{6} x=\left(\sin ^{2} x+\cos ^{2} x\right)^{3}-3 \sin ^{2} x \cos ^{2} x\left(\sin ^{2} x+\cos ^{2} x\right)=1-\frac{3}{4} \sin ^{2}(2 x)
\]

and

\[
\sin ^{4} x+\cos ^{4} x=\left(\sin ^{2} x+\cos ^{2} x\right)^{2}-2 \sin ^{2} x \cos ^{2} x=1-\frac{1}{2} \sin ^{2}(2 x)
\]

the initial equation becomes

\[
1-\frac{3}{4} \sin ^{2}(2 x)=a\left(1-\frac{1}{2} \sin ^{2}(2 x)\right)
\]

from which we immediately get

\[
\sin ^{2}(2 x)=\frac{4(a-1)}{2 a-3}
\]

and due to \(0 \leq \sin ^{2}(2 x) \leq 1\), the condition \(\frac{1}{2} \leq a \leq 1\) for \(a\) follows. Specifically for \(a=\frac{5}{6}\), \(\left({ }^{*}\right)\) is then equivalent to

\[
\sin (2 x)= \pm \frac{\sqrt{2}}{2}
\]

with the obvious solutions

\[
x=(2 k+1) \frac{\pi}{8} \quad(k \in \mathbb{Z})
\]