1. **State the problem:** Solve the equation $$\sin\left(\frac{3}{4} - 4x\right) = -\frac{1}{2}.$$\n\n2. **Recall the solutions for sine:** The sine function equals $$-\frac{1}{2}$$ at angles $$\theta = \frac{7\pi}{6} + 2k\pi$$ and $$\theta = \frac{11\pi}{6} + 2k\pi$$ for any integer $$k$$.\n\n3. **Set the inside of sine equal to these angles:**\n$$\frac{3}{4} - 4x = \frac{7\pi}{6} + 2k\pi$$\nand\n$$\frac{3}{4} - 4x = \frac{11\pi}{6} + 2k\pi$$\nwhere $$k \in \mathbb{Z}$$.\n\n4. **Solve for $$x$$ in each case:**\nFor the first case: $$-4x = \frac{7\pi}{6} + 2k\pi - \frac{3}{4}$$\n$$x = -\frac{1}{4}\left(\frac{7\pi}{6} + 2k\pi - \frac{3}{4}\right)$$\n\nFor the second case: $$-4x = \frac{11\pi}{6} + 2k\pi - \frac{3}{4}$$\n$$x = -\frac{1}{4}\left(\frac{11\pi}{6} + 2k\pi - \frac{3}{4}\right)$$\n\n5. **Simplify expressions:**\nFirst solution: $$x = -\frac{1}{4}\left(\frac{7\pi}{6} - \frac{3}{4} + 2k\pi\right)$$\nSecond solution: $$x = -\frac{1}{4}\left(\frac{11\pi}{6} - \frac{3}{4} + 2k\pi\right)$$\n\n**Final answer:**\n\n$$x = -\frac{1}{4}\left(\frac{7\pi}{6} - \frac{3}{4} + 2k\pi\right)$$ or $$x = -\frac{1}{4}\left(\frac{11\pi}{6} - \frac{3}{4} + 2k\pi\right)$$ where $$k$$ is any integer.