The number of solutions of the equation $\cos \left(x+\frac{\pi }{3}\right)\cos \left(\frac{\pi }{3}-x\right)=\frac{1}{4}{\cos }^{2}2x$, $x\in [-3\pi ,3\pi ]$ is:

$\begin{array}{r}4\left({\cos }^2\left(\frac{\pi }{3}\right)-{\sin }^{2}x\right)={\cos }^{2}2x\\ 4\left(\frac{1}{4}-{\sin }^{2}x\right)={\cos }^{2}2x\\ 1-4{\sin }^{2}x={\cos }^{2}2x\\ 1-2(1-\cos 2x)={\cos }^{2}2x\end{array}$

$$\begin{gathered} \cos 2\mathrm{x}=\mathrm{t} \\ \begin{array}{l}-1+2\cos 2\mathrm{x}={\cos }^{2}2\mathrm{x}\\ {\mathrm{t}}^2-2\mathrm{t}+1=0\\ (\mathrm{t}-1{)}^2=0\\ \mathrm{t}=1\Rightarrow \cos 2\mathrm{x}=1\\ 2\mathrm{x}=2\mathrm{n\pi }\\ \mathrm{x}=\mathrm{n\pi }\\ \mathrm{n}=-3,-2,-1,0,1,2,3\end{array} \end{gathered}$$

The number of solutions of the equation $\cos \left(x+\frac{\pi }{3}\right)\cos \left(\frac{\pi }{3}-x\right)=\frac{1}{4}{\cos }^{2}2x$, $x\in [-3\pi ,3\pi ]$ is: 7