Question

If $\cos 3x+\sin (2x-\frac{7\pi }{6})=-2$, then $x=$
where $k\in z$

• A. $\frac{\pi }{3}(6k+1)$
• B. $\frac{\pi }{3}(6k-1)$
• C. $\frac{\pi }{3}(2k+1)$
• D. None of these

Answer 1

The correct option is A $\frac{\pi }{3}(6k+1)$

$\cos 3x+\sin (2x-\frac{7\pi }{6})=-2$

Since range of $\sin \theta$ and $\cos \theta$ is $[-1,1]$

Therefore,

$\cos 3x=\sin (2x-\frac{7\pi }{6})=-1$

$\therefore \cos 3x=-1$

$\Rightarrow \cos 3x=\cos \pi$

$\Rightarrow x=\frac{\pi }{3}(6k+1)$