Two different trigonometric identities giving two different solutions Using two different sum-difference trigonometric identities gives two different results in a task where the choice of identity seemed unimportant. The task goes as following:

Given  $\cos 2x =-\frac {63}{65} $ , $\cos y=\frac {7} {\sqrt{130}}$, under the condition $0<x,y<\frac {\pi}{2}$, calculate $x+y$.

The first couple of steps are the same: finding  $\sin$ and $\cos$ values for both $x$ and $y$.
From $\cos 2x$ we have:
\begin{align*}
\cos 2x &=-\frac {63}{65} \\
\cos2x &= \cos^2x-\sin^2x = \cos^2x - (1-\cos^2x)=2\cos^2x -1 \\
2\cos^2x -1 &= -\frac {63}{65} \\
2\cos^2x &=\frac {-63+65}{65} \\
\cos^2x &=\frac {1}{65} \\
\cos x &=\frac {1} {\sqrt{65}} 
\end{align*}
(taking only the positive value of $\cos x$ because $\cos x$ is always positive under the given domain)
\begin{align*}
\sin^2x &=1-\frac {1}{65} \\
\sin^2x &=\frac {64}{65}   \\
\sin x &=\frac {8} {\sqrt{65}} 
\end{align*}
(again, only positive value)
From $\cos y$ we have:
\begin{align*}
\cos y &=\frac {7} {\sqrt{130}} \\
\cos^2y &=\frac {49} {130} \\
\sin^2y &=1-\frac {49} {130} \\
\sin^2y &=\frac {81} {130} \\
\sin y &=\frac {9} {\sqrt{130}}
\end{align*}
Now that we've gathered necessary information, we proceed to calculate value of some trigonometric function of $x+y$, hoping we will get some basic angle:
sin(x+y):
\begin{align*}
\sin(x+y) &=\sin x \cos y + \sin y \cos x =\frac {8} {\sqrt{65}} \frac {7} {\sqrt{130}} + \frac {9} {\sqrt{130}}\frac {1} {\sqrt{65}} \\
\sin(x+y) &=\frac {65} {\sqrt{65}\sqrt{130}} \\
\sin(x+y) &=\frac {\sqrt{2}}{2} 
\end{align*}
Thus,
$x+y =\frac {\pi}{4}+2k{\pi}$  OR  $x+y =\frac {3\pi}{4}+2k{\pi}$
Since $x$ and $y$ are in the first quadrant, their sum must lie in first or second quadrant. 
Solutions are:
$x+y= \{\frac {\pi}{4}, \frac {3\pi}{4} \}$ 
cos(x+y):
\begin{align*}
\cos(x+y) &= \cos x \cos y - \sin x \sin y =\frac {1} {\sqrt{65}} \frac {7} {\sqrt{130}} - \frac {8} {\sqrt{65}}\frac {9} {\sqrt{130}} \\
\cos(x+y) &=-\frac {65} {\sqrt{65}\sqrt{130}} \\
\cos(x+y) &=-\frac {\sqrt{2}}{2}
\end{align*}
Thus,
$x+y =\frac {3\pi}{4}+2k{\pi} $   OR   $x+y =\frac {5\pi}{4}+2k{\pi}$
Now we can only have one solution: $x+y=\{\frac {3\pi}{4}\}$
Similar happens with $\cos(x-y)$.
My question is: why do these two formulas give two different solutions? General insight would be great, since I found a lot of examples with similar problems. Thank you in advance.

To know which quadrant $x+y$ is in, you need to know the signs of both $\sin(x+y)$ and $\cos(x+y)$ In your case, $\sin(x+y) \gt 0$ and $\cos(x+y) \lt 0$. This puts $x+y$ in the second quadrant. So the solution is the one angle $x+y = \dfrac{3 \pi}4$.
