Answer
A
Work Step by Step
Comparing $y=-2\sin(4x-3)$ to $y=A\sin(\omega x-\phi)$, note that $A=-2$, $\omega=4$, and $\phi=3$.
Amplitude=$|A|=2$
Period $T=\frac{2\pi}{\omega}=\frac{2\pi}{4}=\frac{\pi}{2}$ and
Phase shift = $\frac{\phi}{\omega}=\frac{3}{4}$
Option A is the correct match.