Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2006 / Paper 1 / 3 / c / Solution

Use the complex exponential forms of the sine and cosine, and put $w=e^{2iz}$. Multiplying the equation by $2e^{iz}$ gives $(2-i)(w-1)/i+(w+1)=0$, which simplifies to $w=1-i$. Taking all branches of the complex logarithm,Thus every solution isThe exponential factor used to multiply the equation is nonzero, so no solutions were introduced or lost.