Question

If $2\sin \theta +1=0$ and $\sqrt{3}\tan \theta =1$ then the most 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

Answer: (C)

$2n\pi +\frac{7\pi }{6}$

$\tan \theta =\frac{1}{\sqrt{3}}$

$\sin \theta =\frac{-1}{2}$

principal solution$=\theta =\frac{7\pi }{6}$

General solution is $\theta =2n\pi +\frac{7\pi }{6}$