We need to evaluate the expression $\frac{\sin(\frac{2\pi}{3}) + \cos(\frac{\pi}{4})}{\sin(\frac{\pi}{6})}$.

1. Problem Description

We need to evaluate the expression $\frac{\sin(\frac{2\pi}{3}) + \cos(\frac{\pi}{4})}{\sin(\frac{\pi}{6})}$.

2. Solution Steps

First, we need to find the values of $\sin(\frac{2\pi}{3})$, $\cos(\frac{\pi}{4})$, and $\sin(\frac{\pi}{6})$.

$\sin(\frac{2\pi}{3})$ can be found using the identity $\sin(\pi - x) = \sin(x)$.

$\sin(\frac{2\pi}{3}) = \sin(\pi - \frac{\pi}{3}) = \sin(\frac{\pi}{3}) = \frac{\sqrt{3}}{2}$.

$\cos(\frac{\pi}{4})$ is a standard value, and $\cos(\frac{\pi}{4}) = \frac{\sqrt{2}}{2}$.

$\sin(\frac{\pi}{6})$ is also a standard value, and $\sin(\frac{\pi}{6}) = \frac{1}{2}$.

Now we substitute these values into the expression:

$\frac{\sin(\frac{2\pi}{3}) + \cos(\frac{\pi}{4})}{\sin(\frac{\pi}{6})} = \frac{\frac{\sqrt{3}}{2} + \frac{\sqrt{2}}{2}}{\frac{1}{2}}$

$= \frac{\frac{\sqrt{3} + \sqrt{2}}{2}}{\frac{1}{2}}$

$= \frac{\sqrt{3} + \sqrt{2}}{2} \cdot \frac{2}{1}$

$= \sqrt{3} + \sqrt{2}$

3. Final Answer

$\sqrt{3} + \sqrt{2}$