Question

If $2\sin \theta +1=0$ and $\sqrt{3}\tan \theta =1$ then find general value of $\theta$ is

• A. $n\pi \pm \frac{\pi }{6}$
• B. $n\pi +{(-1)}^n.\frac{7\pi }{6}$
• C. $2n\pi +\frac{7\pi }{6}$
• D. $2n\pi +\frac{11\pi }{6}$

Answer 1

The correct option is A $n\pi \pm \frac{\pi }{6}$

$2\sin \theta +1=0$ $\tan \theta =\frac{1}{\sqrt{3}}$

$\sin \theta =\frac{-1}{2}$ $\theta =n\pi +\frac{\pi }{6}$

$\theta =n\pi \pm (-1{)}^n\frac{\pi }{6}$

$\theta =n\pi \pm \frac{\pi }{6}$