$\text{ Let }\mathrm{f}:\left[-\frac{\pi }{3},\frac{2\pi }{3}\right]\to [0,4]$ be a function defined by  $\mathrm{f}\left(\mathrm{x}\right)=\sqrt{3}\sin \mathrm{x}-\cos \mathrm{x}+2\text{ then }{\mathrm{f}}^{-1}\left(\mathrm{x}\right)$ equals

$y=2\left(\frac{\sqrt{3}}{2}\sin \left(x\right)-\frac{1}{2}\cos \left(x\right)\right)+2$

$\Rightarrow \frac{\mathrm{y}}{2}-1=\frac{\sqrt{3}}{2}\sin \left(\mathrm{x}\right)-\frac{1}{2}\cos \left(\mathrm{x}\right)$

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

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

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

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

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