Let $\mathrm{f}\left(\mathrm{x}\right)={\sin }^{23}\mathrm{x}-{\cos }^{22}\mathrm{x}\text{ and }\mathrm{g}\left(\mathrm{x}\right)=1+\frac{1}{2}{\tan }^{-1}\left|\mathrm{x}\right|$. Then the number of values of x in the interval $[-10\mathrm{\pi }, 8\mathrm{\pi }]$ satisfying the equation $\mathrm{f}\left(\mathrm{x}\right)=\mathrm{sgn}\left(\mathrm{g}\right(\mathrm{x}\left)\right)$ is _____.

$\mathrm{g}\left(\mathrm{x}\right)=\frac{1}{2}{\tan }^{-1}\left|\mathrm{x}\right|+1$

$\begin{array}{l}\text{ or }\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\mathrm{sgn}\left(\mathrm{g}\right(\mathrm{x}\left)\right)=1\\ \text{ or }\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{\sin }^{23}\mathrm{x}-{\cos }^{22}\mathrm{x}=1\end{array}$

${\text{ or sin}}^{23}\mathrm{x}=1+{\cos }^{22}\mathrm{x}$

which is possible if sin x = 1 and cos x = 0. Therefore,

$\sin \mathrm{x}=1,\mathrm{x}=2\mathrm{n\pi }+\frac{\mathrm{\pi }}{2}$

Hence, $-10\mathrm{\pi }\le 2\mathrm{n\pi }+\frac{\mathrm{\pi }}{2}\le 8\mathrm{\pi } \text{ or }-\frac{21}{4}\le \mathrm{n}\le \frac{15}{4}$

$\text{ or }-5\le \mathrm{n}\le 3$

Hence, number of values of x = 9.