Given that $y = A - B \cos 4x - \frac{1}{2} \sin 2x$ and $\frac{\text{d}^2y}{\text{d}x^2} + 4y = 3 \cos 4x + 1$, find the value of each of the following constants $A$ and $B$.

1. Find the first derivative of $y$ with respect to $x$.

Given $y = A - B \cos 4x - \frac{1}{2} \sin 2x$,
$\frac{\text{d}y}{\text{d}x} = 4B \sin 4x - \cos 2x$

2. Find the second derivative of $y$ with respect to $x$.

$\frac{\text{d}^2y}{\text{d}x^2} = 16B \cos 4x + 2 \sin 2x$

3. Substitute $y$ and $\frac{\text{d}^2y}{\text{d}x^2}$ into the given equation and simplify.

$\frac{\text{d}^2y}{\text{d}x^2} + 4y = (16B \cos 4x + 2 \sin 2x) + 4\left(A - B \cos 4x - \frac{1}{2} \sin 2x\right)$
$= 16B \cos 4x + 2 \sin 2x + 4A - 4B \cos 4x - 2 \sin 2x$
$= 4A + 12B \cos 4x$

4. Compare coefficients to find $A$ and $B$.

We are given that $\frac{\text{d}^2y}{\text{d}x^2} + 4y = 3 \cos 4x + 1$.
Therefore,
$4A + 12B \cos 4x = 1 + 3 \cos 4x$
Comparing the constant terms:
$4A = 1 \implies A = \frac{1}{4}$
Comparing the coefficients of $\cos 4x$:
$12B = 3 \implies B = \frac{1}{4}$