Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2011 / Paper 1 / 7 / Solution

For negative real $x$, the angle $u=\pi e^x$ lies strictly between $0$ and $\pi$. On that interval, $\sin u=1/\sqrt{2}$ has exactly the two solutions $u=\pi /4$ and $u=3\pi /4$. Dividing by $\pi$ and applying the natural logarithm yieldsBoth values are negative, and the restriction $0<u<\pi$ excludes all further periodic sine branches.