The number of $$x \in [0, 2\pi]$$ for which $$\left|\sqrt{2\sin^4 x + 18\cos^2 x} - \sqrt{2\cos^4 x + 18\sin^2 x}\right| = 1$$ is:

$$f(x) = \sqrt{2 \sin^4 x + 18 \cos^2 x} - \sqrt{2 \cos^4 x + 18 \sin^2 x}$$

We need to find the number of values of $$x \in [0, 2\pi]$$ where $$|f(x)| = 1$$. This means $$f(x) = 1$$ or $$f(x) = -1$$.

At $$x = 0$$: $$f(0) = \sqrt{0 + 18(1)} - \sqrt{2(1) + 0} = \sqrt{18} - \sqrt{2} = 3\sqrt{2} - \sqrt{2} = \mathbf{2.82}$$

At $$x = \pi/4$$: Since $$\sin(\pi/4) = \cos(\pi/4)$$, both radicals are equal. $$f(\pi/4) = 0$$

At $$x = \pi/2$$: $$f(\pi/2) = \sqrt{2(1) + 0} - \sqrt{0 + 18(1)} = \sqrt{2} - 3\sqrt{2} = \mathbf{-2.82}$$

By IMVT,  between $$x=0$$ ($$2.82$$) and $$x=\pi/4$$ ($$0$$), the function must pass through $$+1$$ exactly once.

Between $$x=\pi/4$$ ($$0$$) and $$x=\pi/2$$ ($$-2.82$$), the function must pass through $$-1$$ exactly once.

This gives us 2 solutions in the first quadrant.

The terms $$\sin^4 x, \cos^4 x, \sin^2 x,$$ and $$\cos^2 x$$ are all symmetric across all four quadrants.

Total number of solutions $$= 2 \times 4 = \mathbf{8}$$