Clearly, it must be true that $|x|\le 1$ in order to make the term $\sqrt{1-x^2}$ define. Therefore, let $x=\sin\theta$ where $\theta\in[-\frac{\pi}{2}, \frac{\pi}{2}]$. Setting this to the given equation yields \begin{align*} 2\sqrt{2}\sin^2\theta +\sin\theta -\cos\theta -\sqrt{2}&=0\\ \sqrt{2}(2\sin^2\theta-1) +(\sin\theta - \cos\theta)&=0\\ \sqrt{2}(\sin^2\theta - \cos^2\theta) +(\sin\theta - \cos\theta)&=0\\ \sqrt{2}(\sin\theta + \cos\theta)(\sin\theta - \cos\theta) +(\sin\theta - \cos\theta)&=0\\ (\sin\theta-\cos\theta)(\sqrt{2}(\sin\theta+\cos\theta)+1)&=0\\ (\sqrt{2}\sin(\theta- 45^\circ))(2\sin(\theta+45^\circ)+1)&=0 \end{align*} Hence, we find two solutions: \begin{align*} \sin(\theta-45^\circ) =0&\implies \theta = 45^\circ\implies x=\boxed{\frac{\sqrt{2}}{2}}\\ \sin(\theta+45^\circ) =-\frac{1}{2}&\implies\theta=-75^\circ\implies x=\boxed{-\frac{\sqrt{6}+\sqrt{2}}{4}} \end{align*}