We have: $$ x(0)=h\cos(\phi)\cos(\theta) \quad\hbox{and}\quad x(2\pi n)=h\cos(\phi)\cos(\theta)+2\pi na\sin\theta. $$ We want $x(2\pi n)-x(0)=l$, that is $2\pi na\sin\theta=l$, whence: $$ \sin\theta={l\over 2\pi na}. $$
We have: $$ x(0)=h\cos(\phi)\cos(\theta) \quad\hbox{and}\quad x(2\pi n)=h\cos(\phi)\cos(\theta)+2\pi na\sin\theta. $$ We want $x(2\pi n)-x(0)=l$, that is $2\pi na\sin\theta=l$, whence: $$ \sin\theta={l\over 2\pi na}. $$