19: Solving for x you should get
![](gif3/ecb3ce2a.gif)
The problem now is that for a general x the expression
arcsin(sin(x)) does not give x but some x0
from the basic interval of invertibility of sine. If there is some
integer k such that
x − 2kπ
is from this basic interval, then
x0 = x − 2kπ.
Unfortunately, in our case we do not have such k. If we use
x0 = x + 4π,
we get to the basic period
of sine, but not to its first half where we do the inverse. Thus the
expression arcsin(sin(x)) doesn't really give x0
but
x1 = π − x0,
the picture should explain it best.
![](gif3/ecb3ce2b.gif)
Thus we get
![](gif3/ecb3ce2c.gif)
Answer