<s>[INST] Trigonometric equation with high exponents Solve the following equation:
\begin{align}
\sin^{14}x+\cos^{14}x=\frac{169}{64}\cos^{6}2x
\end{align}
I tried setting $\cos^2x=a$ such that the equation would become $a^7+(1-a)^7=\frac{169}{64}(2a-1)^6$ and then expaned both sides, but the obtained polynomial isn't pretty at all...
 [/INST] \begin{align*}
  2\sin^2 x &= 1-\cos 2x \\
  2\cos^2 x &= 1+\cos 2x \\
  128(\sin^{14} x+\cos^{14} x) &= (1-\cos 2x)^7+(1+\cos 2x)^7 \\
  &= 2(1+21\cos^2 2x+35\cos^4 2x+7\cos^6 2x) \\
  169\cos^6 2x &= 1+21\cos^2 2x+35\cos^4 2x+7\cos^6 2x \\
  0 &= 162\cos^6 2x-35\cos^4 2x-21\cos^2 2x-1 \\
  0 &= (2\cos^2 2x-1)(81\cos^4 2x+23\cos^2 2x+1) \\
  \cos^2 2x &= \frac{1}{2} \quad \text{or} \quad
  \frac{-23\pm \sqrt{23^2-4(81)}}{2(81)} < 0
\end{align*}
Can you proceed?
</s>