Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2011 / Paper 1 / 6 / Solution

Put $u=\sin \theta$ and move all terms to one side. The polynomial factors asThe root $u=1$ yields the already supplied $\theta =\pi /2$. In the stated interval, $u=1/2$ gives $\pi /6,5\pi /6$, while $u=-1/2$ gives $-\pi /6,-5\pi /6$. Thus all the other solutions areThere are no endpoint solutions because $\sin (\pm \pi )=0$ is not one of the three roots.