Answer
$0$
$4$
$0$
$-4$
$0$
Work Step by Step
Step 1. Given the function $y=f(x)=4sin(2x-\frac{2\pi}{3})$, we have $f(\frac{\pi}{3})=4sin(2(\frac{\pi}{3})-\frac{2\pi}{3})=4sin(0)=0$
Step 2. Similarly, we have $f(\frac{7\pi}{12})=4sin(2(\frac{7\pi}{12})-\frac{2\pi}{3})=4sin(\frac{\pi}{2})=4$
Step 3. We have $f(\frac{5\pi}{6})=4sin(2(\frac{5\pi}{6})-\frac{2\pi}{3})=4sin(\pi)=0$
Step 4. We have $f(\frac{13\pi}{12})=4sin(2(\frac{13\pi}{12})-\frac{2\pi}{3})=4sin(\frac{3\pi}{2})=-4$
Step 5. Finally, we have $f(\frac{4\pi}{3})=4sin(2(\frac{4\pi}{3})-\frac{2\pi}{3})=4sin(2\pi)=0$
