19: Solving for x you should get

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.

Thus we get

Answer