CAPE Pure Mathematics Unit 1 · May/June 2013 · Paper 1 · Question Q28

The general solution for \sin 2\theta = \sin\frac{\pi}{6} is

1. (A)\theta = \begin{cases} 2n\pi + \frac{\pi}{6} \\ (2n+1)\frac{5\pi}{16} \end{cases}
2. (B)\theta = \begin{cases} n\pi + \frac{\pi}{12} \\ n\pi + \frac{5\pi}{12} \end{cases}
3. (C)\theta = \begin{cases} n\pi + \frac{\pi}{12} \\ (2n\pi)\frac{5\pi}{12} \end{cases}
4. (D)\theta = \begin{cases} n\pi + \frac{\pi}{6} \\ (n+1)\frac{5\pi}{6} \end{cases}