Andre already gave the answer, but let me explain the "generalities".
The main problem is the following:
After you reached the point $\sin(x+y) = \sqrt{2}/2$, you concluded that $x+y$ can take both values $\pi/4$ or $3\pi/4$.
You interpreted this to mean
BOTH $x+y = \pi/4$ and $x+y = 3\pi/4$ are solutions. (That is, the set of solutions is the set $\{\pi/4, 3\pi/4\}$.)
This interpretation is false! The correct interpretation is that
We have ruled out all numbers other than $\pi/4$ and $3\pi/4$ from being possible solutions. (Or, the set of solutions is a subset of the set $\{\pi/4, 3\pi/4\}$.)
There may be other constraints that rule out one (or both) of those values. In your case, knowing that $\cos x \approx 1/8$ should tell you already that $x$ is bigger than $\pi/4$.
Another way to look at this clearly is to look at the directions of implication. From the values of $\sin x$, $\sin y$ etc you can derive the value of $\sin(x+y)$. But just knowing the value of $\sin(x+y)$ you cannot derive the value of $\sin x$, $\sin y$, etc. So the implication going from the "necessary information" step to the "computing $\sin(x+y)$ and $\cos(x+y)$" step is one that loses information.
So that solutions to $\sin(x+y) = \sqrt{2}/2$ are not necessarily solutions to $\cos(x) = 1/\sqrt{65}$ etc.
Consider the equation $x+\sqrt{x}=2$. It can be rearranged as $\sqrt{x}=2-x$, and it becomes $x=4-4x+x^2$ or $x^2-5x+4=0$, so the solutions are $x=1$ or $x=4$.
Oh, wait! If I substitute $x=4$, I get $4+\sqrt{4}=2$, that is, $6=2$. There's something wrong, I guess you agree.
Squaring can introduce spurious roots, in this case also the solution to
$$
-\sqrt{x}=2-x
$$
that is indeed satisfied for $x=4$.
You're doing the same in your second approach.
Note also that $\cos2x=\sqrt{1-\sin^22x}$ is not true in general.
Best Answer
The two solutions are equivalent - they both include all numbers which are congruent to $0$, $\pi/6$, $\pi/2$ or $5\pi/6$ mod $\pi$. It is just that in your first solution you have divided into the first case and a combination of all others, whereas your other solution groups alternate cases together.