Let me reframe this for a bit...also let's assume $\phi_1$ and $\phi_2$ are "interesting" (not $0, \pi$, etc).
Just to be on the same page, let's solve the equation first. Isolating the square roots in both equations gives $q - k_1 \cos\phi_1 = \pm k_2 \sqrt{1-\sin^2 \phi_2}$ and $\mp k_1 \sqrt{1 - \cos^2 \phi_1} = k_2^2 \sin^2 \phi_2$. Squaring and solving gives
$$ \cos \phi_1 = \frac{q^2 + k_1^2 - k_2^2}{2qk_1} $$
which is indeed sign independent. But all of the terms with $\sin \phi_2$ are squared, so $\sin \phi_2$ is not sign independent. Solving the other way, that is, writing $q - k_2 \cos\phi_2 = \pm k_1 \sqrt{1-\sin^2 \phi_1}$ and $\mp k_2 \sqrt{1 - \cos^2 \phi_2} = k_1^2 \sin^2 \phi_1$, gives
$$ \cos \phi_2 = \frac{q^2 + k_2^2 - k_1^2}{2qk_2} $$
(which also makes sense because of symmetry). Similarly, you can't get a value for $\sin \phi_1$ here, since all the $\sin \phi_1$ terms are squared.
The first time, $\cos \phi_1$ was sign-independent and $\sin \phi_2$ wasn't. The second time, $\cos \phi_2$ was sign independent and $\sin \phi_1$ wasn't. Specifically, if you try to solve for the $\sin$s, you get expressions like $\sin \phi = \pm \text{(stuff)}$.
Think of the unit circle for a second. Geometrically, we know the x-coordinates of $\phi_1$ and $\phi_2$, but not their y-coordinates. Though we do know the two possible candidates are $\pm$ of each other. So we are almost done but not quite. We have four possible solutions: $(\phi_1, \phi_2)$, $(\phi_1, -\phi_2)$, $(-\phi_1, \phi_2)$, $(-\phi_1, -\phi_2)$.
Note that $(\phi_1, \phi_2)$ if a solution if and only if $(-\phi_1, -\phi_2)$ is, and the same holds for the two other solution candidates $(\phi_1, -\phi_2)$, $(-\phi_1, \phi_2)$. (See this by plugging them into the second equation.) And $(\phi_1, \phi_2)$ and $(\phi_1, -\phi_2)$ can't both be solutions because the second equation would give $\sin \phi_1 = 0$, a contradiction. Since $(\phi_1, \phi_2)$ is of course a solution to the equation, this tells you that the only two solutions of the equation are $(\phi_1, \phi_2)$ and $(-\phi_1, -\phi_2)$.
This is a long-winded way of saying: You do get solutions that aren't unique when solving this. Specifically, you get 4 solutions, and 2 of them turn out to be fake solutions, leaving you with 2 actual ones. I can't tell where exactly in your question you went wrong, because you didn't explain your solution path, but hopefully this clears things up.
Best Answer
Hint:
$$\begin{align*} \sin(x + y) &= \frac{\sqrt{3}}{2} &\implies x + y = \left\{\frac{\pi}{3},\frac{2\pi}{3}\right\}+2\pi n\\ \sin(x - y) &= \frac{1}{2} &\implies x - y = \left\{\frac{\pi}{6},\frac{5\pi}{6}\right\}+2\pi n\\ \end{align*} $$
Can you take it from here?