trigonometry

Question: The most general value of $\theta $which will satisfy both the equations $\sin \theta =-\frac{1}{2}$and $\tan \theta =\frac{1}{\sqrt{3}}$is

1) $n\pi +{{(-1)}^{n}}\frac{\pi }{6}$

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

3) $2n\pi \pm \frac{\pi }{6}$

4) None of these

Solution:

Explanation: No Explanation

Solution of trigonometric equations

Rate this question:

★ ★ ★ ★ ★

Average rating: (0 votes)