Question

General solution for $\theta$ if $\sin (2\theta +\frac{\pi }{6})+\cos (\theta +\frac{5\pi }{6})=2$ is :

• A. $2n\pi -\frac{\pi }{6}$
• B. $2n\pi +\frac{\pi }{6}$
• C. $2n\pi -\frac{7\pi }{6}$
• D. None of these

Answer 1

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

$\sin (2\theta +\frac{\pi }{6})+\cos (\theta +\frac{5\pi }{6})=2$

only way this is possible is if

$\sin (2\theta +\frac{\pi }{6})=1$ and $\cos (\theta +\frac{5\pi }{6})=$

$\Rightarrow (2\theta +\frac{\pi }{6})=(4m+1)\frac{\pi }{2}$

when $m\in N$

$2\theta =4m\frac{\pi }{2}+(\frac{\pi }{2}-\frac{\pi }{6})$

$2\theta =2m\pi +\frac{\pi }{3}$

$\theta =m\pi +\frac{\pi }{6}$

$\Rightarrow \theta +\frac{5\pi }{6}=2p\pi$

when $p\in N$

$\Rightarrow \theta =2p\pi -\frac{5\pi }{6}$

$\theta =2(p-1)\pi +\frac{7\pi }{6}$

$p\in N$

$\Rightarrow (m-1)\pi +\frac{7\pi }{6}$

$\therefore \theta$ is of the form $2n\pi +\frac{7\pi }{6}$