Given,

$\cos \left(x+\frac{\pi }{3}\right)\cos \left(x-\frac{\pi }{3}\right)=\frac{1}{4}{\cos }^2\left(2x\right)$

$\Rightarrow 2\cos \left(x+\frac{\pi }{3}\right)\cos \left(x-\frac{\pi }{3}\right)=\frac{1}{2}{\cos }^2\left(2x\right)$

$\Rightarrow \cos \left(2x\right)+\cos \left(\frac{2\pi }{3}\right)=\frac{1}{2}{\cos }^{2}2x$

$\Rightarrow \cos 2x+\left(-\frac{1}{2}\right)=\frac{1}{2}{\cos }^{2}2x$

$\Rightarrow {\cos }^{2}2x-2\cos 2x+1=0$

$\Rightarrow {\left(\cos 2x-1\right)}^2=0$or $\cos 2x=1$

$\Rightarrow 2x=-6\pi ,-4\pi ,-2\pi ,0,2\pi ,4\pi ,6\pi$

So, $x\in \left\{-3\pi ,-2\pi ,-\pi ,0,\pi ,2\pi ,3\pi \right\}$

So total $7$ solutions.