 Solution

Again $C=\sin (2\pi /3)=\sqrt{3}/2$, but the principal inverse sine would give $\pi /3$ at $x=1$. The required branch is $v=\pi -\arcsin (Cx)$, corresponding to $k=1$. ThereforeAt $x=1$ the bracket is $2\pi /3$, as required. Both the initial condition and the original differential equation are thus satisfied on the stated branch.