We are asked to calculate $\sin(x + \frac{\pi}{6})$ given that $\sin(x) = \frac{4}{5}$ and $\frac{\pi}{2} \le x \le \pi$.

1. Problem Description

We are asked to calculate $\sin(x + \frac{\pi}{6})$ given that $\sin(x) = \frac{4}{5}$ and $\frac{\pi}{2} \le x \le \pi$.

2. Solution Steps

First, we use the angle sum formula for sine:

$\sin(a+b) = \sin(a) \cos(b) + \cos(a) \sin(b)$

Applying this formula to our problem:

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

We know that $\sin(x) = \frac{4}{5}$, $\cos(\frac{\pi}{6}) = \frac{\sqrt{3}}{2}$, and $\sin(\frac{\pi}{6}) = \frac{1}{2}$. We need to find $\cos(x)$.

We know that $\sin^2(x) + \cos^2(x) = 1$. Therefore, $\cos^2(x) = 1 - \sin^2(x)$.

Substituting $\sin(x) = \frac{4}{5}$, we have:

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

So, $\cos(x) = \pm \sqrt{\frac{9}{25}} = \pm \frac{3}{5}$.

Since $\frac{\pi}{2} \le x \le \pi$, $x$ is in the second quadrant, where cosine is negative. Thus, $\cos(x) = -\frac{3}{5}$.

Now, we substitute all the values into the angle sum formula:

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

3. Final Answer

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