$\cos x+\cos 3x+\cos 5x+\cos 7x=0$, 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
 A: Use product identity $2\cos(x)\cos (y)=\cos(x+y)+\cos(x-y)$
So $\cos x+\cos7x=2\cos4x\cos3x$
And $\cos3x+\cos5x=2\cos4x\cos x$.
Factoring out $2\cos4x$, we get $2\cos4x (\cos3x+\cos x)=0$.
You can solve the 1st factor now right? For the second factor, use the above identity again and you will be done.:)
A: 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
A: $$\text{Using }\cos C+\cos D=2\cos\frac{C+D}2\cos\frac{C-D}2\ \ \ \  (1)$$
$$\cos x+\cos7x=2\cos4x\cos3x\text{ and }\cos3x+\cos5x=2\cos4x\cos x$$
So, the problem reduces to $2\cos4x(\cos3x+\cos x)=0$
If $\displaystyle\cos4x=0, 4x=\frac{(2n+1)\pi}2$ where $n$ is any integer
Else $\cos3x+\cos x=0$
Again, use $(1)$
Generalization:
Using this,
Case $1:$
$$\large S=\sum_{k=0}^{n-1}\cos (a+k \cdot d) =\frac{\sin\left(n \cdot\frac d2\right)}{\sin \frac d2} \times \cos \left( \frac{ 2 a + (n-1)d}2\right)$$
if $\sin \frac d2\ne0\ \ \ \ (*)$
$\iff \frac d2\ne m\pi\iff d\ne2m\pi$ where $m$ is any integer,
If $\displaystyle S=0,$ 
Case $1a):\displaystyle \sin\left(n \cdot\frac d2\right)=0,\implies n \cdot\frac d2=2r\pi\iff d=\frac{4r\pi}n$  where $r$ is any integer
But $n$ must not divide $2r,$ otherwise $d$ will be multiple of $2\pi$ which is prohibited by $(*)$
Case $1b):\cos \left( \frac{ 2 a + (n-1)d}2\right)=0\implies \frac{ 2 a + (n-1)d}2=(2s+1)\frac\pi2\iff 2a+(n-1)d=(2s+1)\pi$  where $s$ is any integer
Case $\displaystyle2:  \sin \frac d2=0 \iff d=2m\pi$ 
$\displaystyle\implies \cos(a+k\cdot d)=\cos(a+k\cdot 2m\pi)=\cos a$ for all integer $k$
$\displaystyle\implies S=\sum_{k=0}^{n-1}\cos (a+k \cdot d) =n\cos a $ which is $0$
As $n\ne0,\displaystyle\implies a=(2t+1)\frac\pi2$  where $t$ is any integer
Observe that in the current problem, $n=4,d=2a$
A: 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 $z=1$.
A: In case you want to crack the question through solving polynomial equation, let $cos(x)=y$
You've found $$64y^7-96y^5+10y^2-4y=0$$
It can be factorized to
$$4y(2y^2-1)(8y^4-8y^2+1)=0$$
Furthermore
$$2y^2-1=(\sqrt{2}y+1)(\sqrt{2}y-1)$$
and
$$(8y^4-8y^2+1)=(2\sqrt{2}y^2+1)^2-(8+4\sqrt{2})y^2=(2\sqrt{2}y^2+\sqrt{8+4\sqrt{2}}y+1)(2\sqrt{2}y^2-\sqrt{8+4\sqrt{2}}y+1)$$
Using quadratic formula you'll get the same solutions other people post.
