For the following two (02) items: Let (2x + 3y) = (1)/(2) and (3x +...

Let $A = 2x + 3y$ and $B = 3x + 2y$.

We are given $\cos A = \dfrac{1}{2}$ with $-\pi < A < \pi$.

The possible values for $A$ are $\dfrac{\pi}{3}$ and $-\dfrac{\pi}{3}$.

We are also given $\cos B = \dfrac{\sqrt{3}}{2}$ with $-\pi < B < \pi$.

The possible values for $B$ are $\dfrac{\pi}{6}$ and $-\dfrac{\pi}{6}$.

We need to find the number of values of $(y - x)$.

Observe that $A - B = (2x + 3y) - (3x + 2y) = y - x$.

The possible values for $A - B$ are:

$\dfrac{\pi}{3} - \dfrac{\pi}{6} = \dfrac{\pi}{6}$

$\dfrac{\pi}{3} - \left(-\dfrac{\pi}{6}\right) = \dfrac{\pi}{2}$

$-\dfrac{\pi}{3} - \dfrac{\pi}{6} = -\dfrac{\pi}{2}$

$-\dfrac{\pi}{3} - \left(-\dfrac{\pi}{6}\right) = -\dfrac{\pi}{6}$

Since all four values are distinct, $(y - x)$ has four values.