The problem asks us to calculate the value of $\sin(x + \frac{\pi}{6})$ given that $\sin(x) = \frac{4}{5}$ and $\frac{\pi}{2} \leq x \leq \pi$.

1. Problem Description

The problem asks us to calculate the value of $\sin(x + \frac{\pi}{6})$ given that $\sin(x) = \frac{4}{5}$ and $\frac{\pi}{2} \leq x \leq \pi$.

2. Solution Steps

First, we will use the sine addition formula:

$$\sin(x + y) = \sin(x)\cos(y) + \cos(x)\sin(y)$$

In our case, $y = \frac{\pi}{6}$, so we have:

$$\sin(x + \frac{\pi}{6}) = \sin(x)\cos(\frac{\pi}{6}) + \cos(x)\sin(\frac{\pi}{6})$$

We are given that $\sin(x) = \frac{4}{5}$. We also know that $\cos(\frac{\pi}{6}) = \frac{\sqrt{3}}{2}$ and $\sin(\frac{\pi}{6}) = \frac{1}{2}$.

So we have:

$$\sin(x + \frac{\pi}{6}) = \frac{4}{5} \cdot \frac{\sqrt{3}}{2} + \cos(x) \cdot \frac{1}{2}$$

We need to find $\cos(x)$. We know that $\sin^2(x) + \cos^2(x) = 1$, so $\cos^2(x) = 1 - \sin^2(x)$.

$$\cos^2(x) = 1 - (\frac{4}{5})^2 = 1 - \frac{16}{25} = \frac{9}{25}$$

Therefore, $\cos(x) = \pm \sqrt{\frac{9}{25}} = \pm \frac{3}{5}$. Since $\frac{\pi}{2} \leq x \leq \pi$, $x$ is in the second quadrant, where cosine is negative. So, $\cos(x) = -\frac{3}{5}$.

Substituting this value into the equation:

$$\sin(x + \frac{\pi}{6}) = \frac{4}{5} \cdot \frac{\sqrt{3}}{2} + (-\frac{3}{5}) \cdot \frac{1}{2} = \frac{4\sqrt{3}}{10} - \frac{3}{10} = \frac{4\sqrt{3} - 3}{10}$$

3. Final Answer

$\sin(x + \frac{\pi}{6}) = \frac{4\sqrt{3} - 3}{10}$