<s>[INST] 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?
 [/INST] When you used the replacement $\cos \alpha = x$, the expression in the bracket became
$$\frac{\cos \alpha}{2} + \frac{\sqrt{3} \sin \alpha}{2} = \cos \frac{\pi}{3} \, \cos \alpha + \sin \frac{\pi}{3} \, \sin \alpha$$
but note that this can be written as both
$$\cos \left( \frac{\pi}{3} - \alpha \right) \quad \text{or} \quad \cos \left( \alpha - \frac{\pi}{3} \right)$$
In order for $\cos^{-1} ( \cos \theta) = \theta,$ we want to avoid $\,\theta \,$ being negative.
When evaluating $f(\frac{2}{3})$, we have that $\,\cos \alpha = \frac{2}{3} \,$, so $\, 0 < \alpha < \frac{\pi}{3}$; but when evaluating $f(\frac{1}{3})$, we have that $\,\cos \alpha = \frac{1}{3} \,$, so $\, 0 < \frac{\pi}{3} < \alpha$.
Thus
$$\begin{align}
f(\frac{2}{3}) &= \alpha + \cos^{-1} \left( \cos \left[ \frac{\pi}{3} - \alpha \right] \right) \\
\\
&= \alpha + \left( \frac{\pi}{3} - \alpha \right) = \boxed{\frac{\pi}{3}}
\end{align}$$
but
$$\begin{align}
f(\frac{1}{3}) &= \alpha + \cos^{-1} \left( \cos \left[ \alpha - \frac{\pi}{3} \right] \right) \\
\\
&= \alpha + \left( \alpha - \frac{\pi}{3} \right) \\
\\
&= 2 \alpha - \frac{\pi}{3} = \boxed{2 \cos^{-1} \left( \frac{1}{3} \right) - \frac{\pi}{3}}
\end{align}$$
</s>