Let \([\cdot ]\) denote the greatest integer function. If the domain of the function \(f\left(x\right)={\cos }^{-1}\left…

For \(f(x)=\cos^{-1}\left(\dfrac{4x+2[x]}{3}\right)\) to be defined, we need \(-1\leq \dfrac{4x+2[x]}{3}\leq 1\).

So \(-3\leq 4x+2[x]\leq 3\).

Let \([x]=n\), where \(n\in \mathbb{Z}\). Then \(n\leq x<n+1\).

The inequality becomes \(-3\leq 4x+2n\leq 3\), hence \(\dfrac{-3-2n}{4}\leq x\leq \dfrac{3-2n}{4}\).

Now check possible integer values of \(n\).

For \(n=-1\), \(-1\leq x<0\) and \(-\dfrac{1}{4}\leq x\leq \dfrac{5}{4}\), giving \(-\dfrac{1}{4}\leq x<0\).

For \(n=0\), \(0\leq x<1\) and \(-\dfrac{3}{4}\leq x\leq \dfrac{3}{4}\), giving \(0\leq x\leq \dfrac{3}{4}\).

No other integer value of \(n\) gives a common interval.

Thus the domain is \(\left[-\dfrac{1}{4},0\right)\cup\left[0,\dfrac{3}{4}\right]=\left[-\dfrac{1}{4},\dfrac{3}{4}\right]\).

So \(\alpha=-\dfrac{1}{4}\) and \(\beta=\dfrac{3}{4}\).

Therefore \(12(\alpha+\beta)=12\left(-\dfrac{1}{4}+\dfrac{3}{4}\right)=12\cdot\dfrac{1}{2}=6\).