$\cos x+\cos 3x+\cos 5x+\cos 7x=0$, Any quick methods? 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

This isn't very fast, but you can use de Moivre's on each term and take the real part like this
    $$\cos(7\theta)  + i\sin(7\theta) = \left[\cos(\theta) + i\sin(\theta)\right]^{7}.$$
By the binomial theorem,
    \begin{align*}
  \left[\cos(\theta) + i\sin(\theta)\right]^{7} = \cos^{7}(\theta) &+ 7\cos^{6}(\theta)i\sin(\theta)+ 21\cos^{5}(\theta)i^{2}\sin^{2}(\theta) + 35\cos^{4}(\theta)i^{3}\sin^{3}(\theta)\\
    &+ 35\cos^{3}(\theta)i^{4}\sin^{4}(\theta) + 21\cos^{2}(\theta)i^{5}\sin^{5}(\theta)+ 7\cos(\theta)i^{6}\sin^{6}(\theta) + i^{7}\sin^{7}(\theta).
 \end{align*}
So $\cos(7\theta)= \cos^{7}(\theta) - 21\cos^{5}(\theta) \sin^{2}(\theta) + 35\cos^{3}(\theta) \sin^{4}(\theta) - 7\cos(\theta)\sin^{6}(\theta)$
and so on
