find the $x$ : $(\cos 3x+\cos 4x)(\cos 3x +\cos x)=\frac{1}{4}$ find the $x$ :
$$(\cos 3x+\cos 4x)(\cos 3x +\cos x)=\frac{1}{4}$$

My try :
$$\cos x +\cos y= 2\cos \left(\frac{x+y}{2}\right) \cos \left(\frac{x-y}{2}\right)$$
So :
$$\left(2\cos \left(\frac{7x}{2}\right)\right) \cos \left(\frac{-x}{2}\right)\left(2\cos \left(\frac{4x}{2}\right)\right) \cos \left(\frac{2x}{2}\right)=\frac{1}{4}$$
$$\cos \left(\frac{7x}{2}\right) \cos \left(\frac{x}{2}\right)\cos (2x) \cos (x)=\frac{1}{16}$$
Now what ?

Substituting $\cos(t) = \dfrac{e^{it} + e^{-it}}{2}$ and letting $y = e^{ix}$, you get:
$$\bigg(\cos(4x) + \cos(3x)\bigg)\bigg(\cos(3x) + \cos(x)\bigg) = \frac{1}{4}$$
$$\bigg(y^4 + y^{-4} + y^3 + y^{-3}\bigg)\bigg(y^3 + y^{-3} + y^1 + y^{-1}\bigg) = 1$$
$$\bigg(\sum_{k = -7}^7 y^k\bigg) + 1 = 1$$
$$\sum_{k = 0}^{14} y^k = 0$$
$$\frac{y^{15} - 1}{y - 1} = 0$$
So $e^{15ix} = 1$ and $e^{ix} \ne 1$, leaving the solutions being $x = \dfrac{2 \pi n}{15}$ for $n \in \mathbb Z$ and $15 \not \mid n$.  I suppose it is just the coincidence of contrived problems that it worked out to be a geometric series.
