The number of solutions of the equation ${\cos }^2\left(\mathrm{x}+\frac{\mathrm{\pi }}{6}\right)+{\cos }^2\mathrm{x}-2\cos \left(\mathrm{x}+\frac{\mathrm{\pi }}{6}\right)\cdot \cos \frac{\mathrm{\pi }}{6}={\sin }^2\frac{\mathrm{\pi }}{6}$in interval $\left(\frac{-\mathrm{\pi }}{2},\frac{\mathrm{\pi }}{2}\right)$ is____.

We have, ${\cos }^2\left(\mathrm{x}+\frac{\mathrm{\pi }}{6}\right)+{\cos }^2\mathrm{x}-2\cos \left(\mathrm{x}+\frac{\mathrm{\pi }}{6}\right)\cdot \cos \frac{\mathrm{\pi }}{6}={\sin }^2\frac{\mathrm{\pi }}{6}$

    ${\left(\cos \left(\mathrm{x}+\frac{\mathrm{\pi }}{6}\right)-\cos \left(\frac{\mathrm{\pi }}{6}\right)\right)}^2={\sin }^2\mathrm{x}$
$\begin{array}{l}\Rightarrow 4{\sin }^2\left(\frac{\mathrm{x}}{2}\right){\sin }^2\left(\frac{\mathrm{x}}{2}+\frac{\mathrm{\pi }}{6}\right)-4{\sin }^2\left(\frac{\mathrm{x}}{2}\right)\cdot {\cos }^2\left(\frac{\mathrm{x}}{2}\right)=0\\ \therefore \sin \frac{\mathrm{x}}{2}=0\text{ or }{\sin }^2\left(\frac{\mathrm{x}}{2}+\frac{\mathrm{\pi }}{6}\right)={\cos }^2\frac{\mathrm{x}}{2}\\ \Rightarrow \mathrm{x}=2\mathrm{m\pi },\mathrm{m}\in \mathrm{Z}\text{ or }\frac{\mathrm{\pi }}{2}-\left(\frac{\mathrm{x}}{2}+\frac{\mathrm{\pi }}{6}\right)=\mathrm{n\pi }\pm \frac{\mathrm{x}}{2},\mathrm{n}\in \mathrm{I}\\ \therefore \mathrm{x}=0\text{ or }\mathrm{x}=\frac{\mathrm{\pi }}{3}\end{array}$