We are given that $\sin{\alpha} = \frac{4}{5}$ and $\frac{\pi}{2} \le \alpha < \pi$. We need to find the value of $\sin(x + \frac{\pi}{6})$. The original problem likely intends $\alpha$ to be $x$, but in that case $\sin(x+\frac{\pi}{6})$ cannot be simplified to a single value, as $x$ is not uniquely determined. We will assume the problem intends to evaluate $\sin(\alpha + \frac{\pi}{6})$ instead.

1. Problem Description

We are given that $\sin{\alpha} = \frac{4}{5}$ and $\frac{\pi}{2} \le \alpha < \pi$.

We need to find the value of $\sin(x + \frac{\pi}{6})$. The original problem likely intends $\alpha$ to be $x$, but in that case $\sin(x+\frac{\pi}{6})$ cannot be simplified to a single value, as $x$ is not uniquely determined. We will assume the problem intends to evaluate $\sin(\alpha + \frac{\pi}{6})$ instead.

2. Solution Steps

First, since $\sin{\alpha} = \frac{4}{5}$ and $\frac{\pi}{2} \le \alpha < \pi$, $\alpha$ lies in the second quadrant. We can find $\cos{\alpha}$ using the Pythagorean identity:

$\sin^2{\alpha} + \cos^2{\alpha} = 1$

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

Since $\alpha$ is in the second quadrant, $\cos{\alpha}$ is negative.

$\cos{\alpha} = -\sqrt{\frac{9}{25}} = -\frac{3}{5}$.

Now we need to find $\sin(\alpha + \frac{\pi}{6})$. Using the sine addition formula, we have:

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

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

We know that $\sin{\alpha} = \frac{4}{5}$, $\cos{\alpha} = -\frac{3}{5}$, $\cos{\frac{\pi}{6}} = \frac{\sqrt{3}}{2}$, and $\sin{\frac{\pi}{6}} = \frac{1}{2}$.

Substituting these values into the formula, we get:

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

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

3. Final Answer

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