Given that the function $\text{f}(x) = 1 + 2\sin^2 3x + 4\cos^2 3x$,

1. Simplify the expression using the Pythagorean identity $\sin^2 A + \cos^2 A = 1$.

$\begin{aligned} \text{f}(x) &= 1 + 2\sin^2 3x + 4\cos^2 3x \\ &= 1 + 2(\sin^2 3x + \cos^2 3x) + 2\cos^2 3x \\ &= 1 + 2(1) + 2\cos^2 3x \\ &= 3 + 2\cos^2 3x \end{aligned}$

2. Apply the double angle formula to express in terms of $\cos 6x$.

Using $\cos 2A = 2\cos^2 A - 1$, we have $2\cos^2 3x = \cos 6x + 1$.
$\begin{aligned} \text{f}(x) &= 3 + (\cos 6x + 1) \\ &= 4 + \cos 6x \end{aligned}$

1. Express $\text{f}(x)$ in the form $a + b\cos 6x$ (from part i)

$\text{f}(x) = 1 + 2\sin^2 3x + 4\cos^2 3x$
$= 1 + 2(\sin^2 3x + \cos^2 3x) + 2\cos^2 3x$
$= 1 + 2(1) + (\cos 6x + 1)$
$= 4 + \cos 6x$

2. Set up the equation $\text{f}(x) = 3.5$ and find the basic angle

$4 + \cos 6x = 3.5$
$\cos 6x = -0.5$
Basic angle $\alpha = \cos^{-1}(0.5) = \frac{\pi}{3}$

3. Find all possible values of $x$ in the given domain

Given $0 \le x \le \frac{\pi}{2}$, the domain for $6x$ is $0 \le 6x \le 3\pi$.
Since $\cos 6x$ is negative, $6x$ lies in the 2nd and 3rd quadrants.
$6x = \pi - \frac{\pi}{3}, \pi + \frac{\pi}{3}, 3\pi - \frac{\pi}{3}$
$6x = \frac{2\pi}{3}, \frac{4\pi}{3}, \frac{8\pi}{3}$
$x = \frac{\pi}{9}, \frac{2\pi}{9}, \frac{4\pi}{9}$