Question

Question inside! :)

Answer 1

what do u think @terenzreignz ? :)

Answer 2

It won't be a function if there was more than one :)
Good job.

Answer 3

Wait... this isn't a function, is it :D

Answer 4

lol :P

wait so is the answer still 1 then? :)

Answer 5

Nope.

Answer 6

aww darn :( then how can i find the right one @terenzreignz ? :D

Answer 7

wait so if it isn't a function, then what is it? :/

Answer 8

there are infinite solutions

\[\cos (\pi/2)=0\\
\cos(3\pi/2)=0\\
\vdots\\
\cos\left[(2n+1){\pi\over2}\right]=0
\]

Answer 9

Because these functions (sines and cosines) aren't one-to-one...
For instance, \(\large \frac{\pi}{2}\) a solution, but also \(\large \frac{3\pi}{2}\)

and many others ;)

Answer 10

every time you trace the angle to these veritical positions, cosine ratio becomes "0"