Problem 931

Let $z = -x - \frac{\pi}{6}$. Then $z \in [\frac{\pi}{6}, \frac{\pi}{4}]$, and $2z \in [\frac{\pi}{3}, \frac{\pi}{2}]$. We have
$$\tan\left(x + \frac{2\pi}{3}\right) = -\cot\left(x + \frac{6}{\pi}\right) = \cot z.$$
Then
$$y = \cot z + \tan z + \cos z = \frac{2}{\sin 2z} + \cos z.$$
Since both $\frac{2}{\sin 2z}$ and $\cos z$ are monotonic decreasing in this case, so $y$ reaches the maximum at $z = \frac{\pi}{6}$, where $y_{\max} = \frac{2}{\sin \frac{\pi}{3}} + \cos \frac{\pi}{6} = \frac{4}{\sqrt{3}} + \frac{\sqrt{3}}{2} = \frac{11}{6}\sqrt{3}$.

Answer: C.