Solve $\cos (2x+3y)=\frac{1}{2}$ and $\cos (3x+2y)=\frac{\sqrt{3}}{2}$ We have to solve for $x$ and $y$:
$$\left \{
 \begin{align*}
   \cos (2x+3y) &= \frac{1}{2} \\
   \cos (3x+2y) &= \frac{\sqrt{3}}{2}
  \end{align*}
\right.$$
I got
$$\left \{
 \begin{align*}
   2x+3y &= 2n\pi  \pm\frac{\pi}{3} \\
   3x+2y &= 2m\pi \pm\frac{\pi}{6}
  \end{align*}
\right.$$
The final answer in the book is 
$x=\dfrac{1}{5}
\left[ (6m-4n)\pi \pm \dfrac{\pi}{2} \mp \dfrac{2\pi}{3} \right]$
Why is it $\mp \dfrac{2\pi}{3}$?
If for some value we had taken
$$2x+3y=2n\pi -\frac{\pi}{3}$$
and
$$3x+2y=2m\pi +\frac{\pi}{6}$$
The value of $x$ would be
$$\frac{1}{5}[(6m-4n)\pi+\frac{\pi}{2}+\frac{2\pi}{3}]$$
which is not included in the general solution.

$1)$ You have:
$$2x+3y = 2n\pi  \pm\frac{\pi}{3}$$
Multiply by $-2$ and get:
$$-4x-6y = -4n\pi  \mp\frac{2\pi}{3} \quad (I)$$
$2)$ You have:
$$ 3x+2y= 2m\pi \pm\frac{\pi}{6}$$
Multiply by $3$ and get:
$$9x+6y= 6m\pi \pm \frac{\pi}{2}\quad (II)$$
$(I)+(II)$:
$$5x=(6m-4n)\pi \mp\frac{2\pi}{3}\pm \frac{\pi}{2}$$
PS: $\mp\frac{2\pi}{3}$ means that you got opposite signals when multiplied by a negative number, in that case it was $-2$. 
