1. The problem is to solve the equation $\sin\left(\frac{3}{4} - 4x\right) = -\frac{1}{2}$.\n\n2. Recall that $\sin\theta = -\frac{1}{2}$ 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 the sine equal to these angles:\n$$\frac{3}{4} - 4x = \frac{7\pi}{6} + 2k\pi$$\n$$\frac{3}{4} - 4x = \frac{11\pi}{6} + 2k\pi$$\nwhere $k$ is any integer.\n\n4. Solve each for $x$:\nFor the first equation:\n$$ -4x = \frac{7\pi}{6} + 2k\pi - \frac{3}{4} $$\n$$ x = -\frac{1}{4} \left(\frac{7\pi}{6} + 2k\pi - \frac{3}{4} \right) $$\n\nFor the second equation:\n$$ -4x = \frac{11\pi}{6} + 2k\pi - \frac{3}{4} $$\n$$ x = -\frac{1}{4} \left(\frac{11\pi}{6} + 2k\pi - \frac{3}{4} \right) $$\n\n5. These expressions represent the general solution for $x$, where $k$ is any integer.\n\nFinal answer:\n$$ x = -\frac{1}{4} \left( \frac{7\pi}{6} + 2k\pi - \frac{3}{4} \right), \quad x = -\frac{1}{4} \left( \frac{11\pi}{6} + 2k\pi - \frac{3}{4} \right), \quad \text{for } k\in\mathbb{Z}. $$