Determining $A+B$, given $\sin A + \sin B = \sqrt{\frac{3}{2}}$ and $\cos A - \cos B = \sqrt{\frac12}$. Different approaches give different answers. The question:

Determine $A + B$ if $A$ and $B$ are acute angles such that:
$$\sin A + \sin B = \sqrt{\frac{3}{2}}$$
$$\cos A - \cos B = \sqrt{\frac{1}{2}} $$

Here are the two solutions that I found:

I think the problem with the second solution may have to do with the assumption that:
$$\cos\left(A + \frac{\pi}{6}\right) = \cos\left(B - \frac{\pi}{6}\right)$$
is equivalent to
$$A + \frac{\pi}{6} = -B + \frac{\pi}{6}$$
But can't you say that $\cos N = \cos M$ is equivalent to $\pm N = \pm M$ for all values of $N$ and $M$ (because cosine is an even function)?

Your first way is right.
In the second solution $$A+\frac{\pi}{6}=B-\frac{\pi}{6},$$ which is very well or $$A+B=0,$$ which is impossible. 
Another way:
$$2\sin\frac{A+B}{2}\cos\frac{A-B}{2}=\sqrt{\frac{3}{2}}$$ and
$$2\sin\frac{A+B}{2}\sin\frac{B-A}{2}=\sqrt{\frac{1}{2}}.$$
Thus,
$$\tan\frac{B-A}{2}=\frac{1}{\sqrt3}$$ and since $A$ and $B$ they are acute angles, we obtain $$B-A=60^{\circ},$$
$$\sin\frac{A+B}{2}=\frac{1}{\sqrt2}$$ and
$$A+B=90^{\circ}.$$
