Calculus
y=3cos(2x−π2)+2y=3\(\cos\[\left\)(2x-\(\frac{\pi}{2}\]\right\))+2
y=3cos(2x−π2)−2y=3\(\cos\[\left\)(2x-\(\frac{\pi}{2}\]\right\))-2
y=3cos(2x+π2)+2y=3\(\cos\[\left\)(2x+\(\frac{\pi}{2}\]\right\))+2
y=3cos(2x+π2)−2y=3\(\cos\[\left\)(2x+\(\frac{\pi}{2}\]\right\))-2