Find arccos[cos(7pi/2)] show work. i will mark brainliest

Arccos(cos 7pi/2) 
cos 7pi/2 = cos 630 = cos (630-360) = cos 270 = 0 
arccos(0) = 90 since cos 90 = cos 270 
(x):0 -30 -45 -60 -90 
sin (x)0 - 1/2 - 1/2 √3 - 1/2 √3 - 1 
cos (x)1 - 1/2 √3 - 1/2 √3 - 1/2 - 0 
tan (x)0 - 1/√3 - 1 - √3 - ∞ 

Rules for sin cos tan value in each quadrant : 
quadrant = I(0-90)- II (90-180)- III(180-270) - IV(270-360) 
sin (x) = positive- positive - negative - negative 
cos (x) = positive- negative- negative - positive 
tan (x) = positive- negative- positive - negative

I’m pretty sure the answer is a