Let $f:\left[-\frac{π}{3},\frac{2π}{3}\right] → [0, 4]$ be a function defined as $f(x) = \sqrt{3}\sin x − \cos x + 2$. Then $f^{−1}(x)$ is given by

$\frac{2π}{3}-\cos^{-1}\left(\frac{x-2}{3}\right)$

$f(x) = \sqrt{3}\sin x − \cos x + 2=2\sin\left(x-\frac{π}{6}\right)+2$

Since f(x) is one-one and onto, f is invertible.

$⇒f^{-1}(x)=\sin^{-1}\left(\frac{x}{2}-1\right)+\frac{π}{6}$. Because $\left|\frac{x}{2}-1\right|≤1$ for all x ∈ [0, 4]

Also using $\sin^{-1}α+\cos^{-1}α=\frac{π}{2}$

$f^{-1}(x)=\frac{π}{2}-\cos^{-1}\left(\frac{x-2}{2}\right)+\frac{π}{6}=\frac{2π}{3}-\cos^{-1}\left(\frac{x-2}{3}\right)$