Simplifying $\cos^{-1}x +\cos^{-1}\left(\frac{x}{2} + \frac{\sqrt{3-3x^2}}{2}\right)$ A question has this equation: $$f(x) = \cos^{-1}x + \cos^{-1}\left(\frac{x}{2} + \frac{\sqrt{3-3x^2}}{2}\right)$$ and you're supposed to simplify it and find $f\left(\frac{2}{3}\right)$ and $f\left(\frac{1}{3}\right)$.
By taking $\cos\alpha  = x$, the equation on the right can be simplified to $\cos^{-1}\left(\cos\left(\frac {\pi}{3} - \alpha\right)\right)$. Finally, you get $\frac{\pi}{3}$ as the final answer.
But the answers are $\frac{\pi}{3}$ and  $2\cos^{-1}\left(\frac{1}{3}\right)-\frac{\pi}{3}$. How does it work out that way?

$$
f(\cos\alpha)=\cos^{-1}(\cos\alpha)+\cos^{-1}\bigg(\frac{1}2\cos\alpha+\frac{\sqrt3}2\sin\alpha\bigg)\\
= \cos^{-1}(\cos\alpha) +\cos^{-1}\bigg(\cos\frac \pi 3\cos\alpha+\sin\frac{\pi}3\sin\alpha\bigg)\\
=\cos^{-1}(\cos\alpha)+\cos^{-1}\bigg[\cos\bigg(\frac\pi3-\alpha\bigg)\bigg]\\
= \pm\alpha+2\pi k_1 \pm\ (\frac\pi3-\alpha)+2\pi k_2 \\
=\pm\alpha  \pm\ (\frac\pi3-\alpha)+2\pi k
$$
where $k,k_1,k_2 $ are integers.
For the first case we get the prescribed solution by taking the $+$ signs in both and $k=0$. In the second case, if we take the $+$ sign on the first and the $-$ sign on the second, and $k=0$ we get:
$$
2\alpha-\frac\pi3=2\cos^{-1}\big(\frac13 \big)-\frac\pi3
$$
