Question

What is most general value of $\theta$ which satisfies both the equations.
$\sin \theta =-1/2$ and $\tan \theta =1/\sqrt{3}$.

Answer 1

We shall first consider values of $\theta$ between $0$ and $2\pi$.
$\sin \theta =-\frac{1}{2}=-\sin \frac{\pi }{6}=\sin (\pi +\frac{\pi }{6})$
or $\sin (2\pi -\pi /6)$
$\therefore \theta =7\pi /6;11\pi /6$
$\tan \theta =1/\sqrt{3}=\tan \left(\pi /6\right)=\tan (\pi +\pi /6)$
$\therefore \theta =\pi /6,7\pi /6$.
Hence the value of $\theta$ between $0$ and $2\pi$ which satisfies both the equations is $7\pi /6$
Hence the general value of $\theta$ is $2n\pi +7\pi /6$, where $n\in I$.