How to solve the following equation by a quick method?
\begin{eqnarray}
\\\cos x+\cos 3x+\cos 5x+\cos 7x=0\\
\end{eqnarray}
If I normally solve the equation, it takes so long time for me.
I have typed it into a solution generator to see the steps. One of the steps shows that :
\begin{eqnarray}
\\\cos x+\cos 3x+\cos 5x+\cos 7x&=&0\\
\\-4\cos x+40\cos ^3x-96\cos ^5x+64\cos ^7x&=&0\\
\end{eqnarray}
How can I obtain this form? It seems very quick. or this quick methd do not exist?
Thank you for your attention
Best Answer
Write $e^{i\theta}+e^{-i\theta}=2\cos \theta$. Then
$ 0=2(\cos \theta+\cos 3\theta+\cos 5\theta+\cos 7\theta)=e^{i\theta}+e^{-i\theta}+e^{-i3\theta}+e^{i3\theta}+e^{-i5\theta}+e^{i5\theta}+e^{-i7\theta}+e^{i7\theta} $
Multiply by $e^{i7\theta}$:
$ 0=e^{i8\theta}+e^{i6\theta}+e^{i4\theta}+e^{i10\theta}+e^{i2\theta}+e^{i12\theta}+e^{-i0\theta}+e^{i14\theta} $
Set $z=e^{i2\theta}$:
$ 0=1+z+z^2+z^3+z^4+z^5+z^6+z^7=\dfrac{1-z^8}{1-z} $
So, $z^8=1$ and $e^{i16\theta}=1$. This means that $\theta=\dfrac{k\pi}{8}$, for $k\in\mathbb Z$, $k$ not a multiple of $8$, because we need to avoid $z=1$.