1. The problem asks for the domain of definition of the function $\arcsin(3 - 4x)$.\n\n2. Recall that the domain of $\arcsin(y)$ is $-1 \leq y \leq 1$. This means the expression inside the arcsin, here $3 - 4x$, must satisfy:\n$$-1 \leq 3 - 4x \leq 1$$\n\n3. Solve the inequalities step-by-step:\n\nFirst inequality: $$-1 \leq 3 - 4x$$\nSubtract 3 from both sides:\n$$-1 - 3 \leq 3 - 4x - 3$$\n$$-4 \leq -4x$$\nDivide both sides by $-4$ (remember to reverse inequality sign when dividing by negative):\n$$\cancel{-4} \div \cancel{-4} \geq \cancel{-4x} \div \cancel{-4}$$\n$$1 \geq x$$\n\nSecond inequality: $$3 - 4x \leq 1$$\nSubtract 3 from both sides:\n$$3 - 4x - 3 \leq 1 - 3$$\n$$-4x \leq -2$$\nDivide both sides by $-4$ (reverse inequality):\n$$\cancel{-4x} \div \cancel{-4} \geq \cancel{-2} \div \cancel{-4}$$\n$$x \geq \frac{1}{2}$$\n\n4. Combine both results to get the domain:\n$$\frac{1}{2} \leq x \leq 1$$\n\n5. Therefore, the domain of definition of $\arcsin(3 - 4x)$ is $\boxed{\frac{1}{2} \leq x \leq 1}$, which corresponds to option C.