18: Solving for y you should get
The problem now is that for a general x the expression
arccos(cos(x)) does not give x but some x0
from the basic interval of invertibility of cosine. If there is some
integer k such that
x − 2kπ
is from this basic interval, then
x0 = x − 2kπ,
the picture should explain it best.
In our case
x − 6π
works, so
Answer