Let $[\cdot]$ denote the greatest integer function. If the domain of the function $f(x) = \cos^{-1}\left(\dfrac{4x+2[x]}{3}\right)$ is $[\alpha, \beta]$, then $12(\alpha + \beta)$ is equal to:

For the domain of $f(x) = \cos^{-1}\left(\dfrac{4x+2[x]}{3}\right)$, the argument must satisfy:

$-1 \le \dfrac{4x+2[x]}{3} \le 1$

$-3 \le 4x + 2[x] \le 3$

Let $x = I + f$, where $I = [x] \in \mathbb{Z}$ and $f = \{x\} \in [0, 1)$. Substituting $x = I + f$ into the inequality:

$-3 \le 4(I + f) + 2I \le 3$

$-3 \le 6I + 4f \le 3$

Since $0 \le f < 1$, we have $0 \le 4f < 4$. We can check the possible integer values of $I$:
If $I \ge 1$, $6I + 4f \ge 6(1) + 0 = 6 > 3$, which gives no solution.
If $I \le -2$, $6I + 4f < 6(-2) + 4 = -8 < -3$, which gives no solution.

Thus, the only possible values for $I$ are $0$ and $-1$.

Case 1: $I = 0$
$-3 \le 6(0) + 4f \le 3 \Rightarrow -3 \le 4f \le 3 \Rightarrow -\dfrac{3}{4} \le f \le \dfrac{3}{4}$
Since $f \in [0, 1)$, we get $0 \le f \le \dfrac{3}{4}$.
Therefore, $x = I + f \in \left[0, \dfrac{3}{4}\right]$.

Case 2: $I = -1$
$-3 \le 6(-1) + 4f \le 3 \Rightarrow 3 \le 4f \le 9 \Rightarrow \dfrac{3}{4} \le f \le \dfrac{9}{4}$
Since $f \in [0, 1)$, we get $\dfrac{3}{4} \le f < 1$.
Therefore, $x = I + f \in \left[-1 + \dfrac{3}{4}, -1 + 1\right) = \left[-\dfrac{1}{4}, 0\right)$.

Taking the union of the intervals from both cases, the domain is $x \in \left[-\dfrac{1}{4}, \dfrac{3}{4}\right]$.

Comparing this with $[\alpha, \beta]$, we get $\alpha = -\dfrac{1}{4}$ and $\beta = \dfrac{3}{4}$.

Finally, evaluating the required expression:

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

Answer: $6$