Given $\cos (\theta)$ and $\sin (\theta)$, find $2\theta$ I am working on my scholarship exam. I worked through almost final step but got my answer wrong. Could you please have a look?

If $\cos (\theta) = \sqrt{\frac{1}{2}+\frac{1}{2\sqrt{2}}}$ and $\sin
(\theta) = -\sqrt{\frac{1}{2}-\frac{1}{2\sqrt{2}}}$ with
   $0\leq\theta<2\pi$, it follows that $2\theta = ..... \pi$

What I have got is below:
$\sin(2\theta)=2\sin\theta\cos\theta$
Then, $\sin(2\theta)=-\frac{1}{\sqrt{2}}$
Hence, $2\theta = \frac{5\pi}{4}$ or $\frac{7\pi}{4}$ (quadrant 3 or 4)
$\theta=\frac{5\pi}{8}$ or $\frac{7\pi}{8}$
Since $\cos\theta$ is positive and $\sin\theta$ is negative, $\theta$ should be in quadrant 4 but my  $\theta$'s are not. So I cannot use my $2\theta$ as a final answer.
However, the answer key provided is $\theta=\frac{15\pi}{4}$, Why do you think that is the case? How did they get to this answer? Please help.

If $0\le \theta<2\pi$, then $0\le 2\theta<4\pi$. You should have $4$ possible values of $\theta$.
$\cos (\theta) = \sqrt{\frac{1}{2}+\frac{1}{2\sqrt{2}}}$ and $\sin
(\theta) = -\sqrt{\frac{1}{2}-\frac{1}{2\sqrt{2}}}$ imply that $\sin(2\theta)=-\dfrac1{\sqrt2}$, but not the other way round.

Note that $\displaystyle \tan\theta=\frac{-\sqrt{\frac{1}{2}-\frac{1}{2\sqrt{2}}}}{\sqrt{\frac{1}{2}+\frac{1}{2\sqrt{2}}}}=-\sqrt{\frac{\sqrt2-1}{\sqrt2+1}}=1-\sqrt{2}$.
So $\theta=n\pi-\dfrac{\pi}{8}$, ($n\in\mathbb{Z}$).
[Note: $\frac{\tan(-\frac\pi8)}{1-\tan^2(-\frac\pi8)}=\tan2(-\frac\pi8)=-1$ implies that $\tan(-\frac\pi8)=1-\sqrt2$.]
As $\theta\in[0,2\pi)$, $\cos\theta>0$ and $\sin\theta<0$, we have $\theta=2\pi-\dfrac{\pi}{8}$ and hence $2\theta=\dfrac{15\pi}4$.
