<s>[INST] 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.
 [/INST] To get rid of $y$,
the first equation
is multiplied by 2
and the second by 3
giving
$\pm 2\pi/3\pm\pi/2$.
To get rid of $x$,the first equation is multiplied by 3 
and the second by 2
giving
$\pm \pi\pm\pi/3$.
</s>