Question

If $2\sin \theta -1=0$ and $2\cos \theta +\sqrt{3}=0$ then the general values of $\theta$ are given by

• A. $2n\pi +\frac{\pi }{6}$
• B. $2n\pi +\frac{\pi }{4}$
• C. $2n\pi +\frac{7\pi }{6}$
• D. $2n\pi +\frac{5\pi }{6}$

Answer 1

The correct option is D $2n\pi +\frac{5\pi }{6}$

$2sin\theta -1=0$

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

$cos\theta =cos\frac{\pi }{6}$

$\theta =2n\pi \pm \frac{\pi }{6}$

$2cos\theta +\sqrt{3}=0$

$cos\theta =\frac{-\sqrt{3}}{2}=cos\frac{5\pi }{6}$

$\theta =2n\pi \pm \frac{5\pi }{6}$

If you take LCM of both the solutions,
you get,
$\theta =2n\pi \pm \frac{5\pi }{6}$