So I'm reading Landau&Lifshitz's book on quantum mechanics and there's this solution to the problem of energy levels in a non-symmetrical potential well. All is good but I don't quite get where the $\pi n$ comes from after eliminating $\delta$. Isn't the function $\arcsin[\sin(x)]$ supposed to change in sign after $x=\pi$? For a bit more context here are the two lines that made me puzzled: $\sin(\delta)=\frac{k\hbar}{\sqrt{2mU_1}},\sin(ka+\delta)=\!-\!\frac{k\hbar}{\sqrt{2mU_2}}\!\implies\! ka=\pi n-\arcsin\left(\frac{k\hbar}{\sqrt{2mU_1}}\right)-\arcsin\left(\frac{k\hbar}{\sqrt{2mU_2}}\right)$
To be fair, he mentions that n=1,2,3... and that the values of $\arcsin$ are taken between $0$ and $\frac{\pi}{2}$, but I still don't quite get how that explains everything.