20: Solving for x you should get

The problem now is that for a general z the expression arccos(cos(z)) does not give z but some z₀ from the basic interval of invertibility of cosine. If there is some integer k such that z − 2kπ is from this basic interval, then z₀ = z − 2kπ.

In our case we have to first observe that z = x/3 is from the interval [5π,6π]. This means that we do not have k as described above. The best we can do it to say that z₀ = z − 4π lies in the basic period of cosine, but not in its first half where we do the inverse. Thus the expression arccos(cos(z)) doesn't really give z₀ but z₁ = 2π − z₀, the picture should explain it best.

Thus we get

Answer