Solution:
$1-\sin^{4} x-\cos^{2} x = \frac{1}{16} \Rightarrow (16-16 \cos^{2} x) - \sin^{4} x - 1 = 0 \Rightarrow 16 \sin^{4} x - 16 \sin^{2} x + 1 = 0$.

Use the quadratic formula in $\sin x$ to obtain $\sin^{2} x = \frac{1}{2} \pm \frac{\sqrt{3}}{4}$.

Since $\cos 2x = 1 - 2 \sin^{2} x = \pm \frac{\sqrt{3}}{2}$, we get $x = \pm \frac{\pi}{12}, \pm \frac{5\pi}{12}$.