1. Stating the problem: Solve the equation $$8 \sin^2(x) \cos^2(x) = 1$$. 2. Use the double-angle identity for sine: $$\sin(2x) = 2 \sin(x) \cos(x)$$, so $$\sin^2(2x) = 4 \sin^2(x) \cos^2(x)$$. 3. Rewrite the given equation in terms of $$\sin^2(2x)$$: $$8 \sin^2(x) \cos^2(x) = 1 \Rightarrow 2 \times 4 \sin^2(x) \cos^2(x) = 1 \Rightarrow 2 \sin^2(2x) = 1$$. 4. Simplify: $$2 \sin^2(2x) = 1 \Rightarrow \sin^2(2x) = \frac{1}{2}$$. 5. Take the square root: $$\sin(2x) = \pm \frac{1}{\sqrt{2}} = \pm \frac{\sqrt{2}}{2}$$. 6. Solve for $$2x$$: $$2x = \arcsin\left( \pm \frac{\sqrt{2}}{2} \right) + k\pi, \quad k \in \mathbb{Z}$$. This corresponds to angles: $$2x = \frac{\pi}{4} + k\pi \quad \text{or} \quad 2x = \frac{3\pi}{4} + k\pi$$. 7. Divide both sides by 2 to solve for $$x$$: $$x = \frac{\pi}{8} + \frac{k\pi}{2} \quad \text{or} \quad x = \frac{3\pi}{8} + \frac{k\pi}{2}, \quad k \in \mathbb{Z}$$. Final answer: $$x = \frac{\pi}{8} + \frac{k\pi}{2} \quad \text{or} \quad x = \frac{3\pi}{8} + \frac{k\pi}{2}, \quad k \in \mathbb{Z}$$.