Question

If $\cos 3x+\sin (2x-\frac{7\pi }{6})=-2$ then $x$ is equal to $\left(kϵz\right)$

• 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 D None of these

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

is only possible when

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

$\Rightarrow 3x=(2n+1)\pi$ and $2\pi -\frac{7\pi }{6}=(4n-1)\frac{\pi }{2}$

$\Rightarrow x=(2n+1)\frac{\pi }{3}$ and $2x=2n\pi -\frac{\pi }{2}+\frac{7\pi }{6}$

$\Rightarrow x=(2n+1)\frac{\pi }{3}$ and $x=n\pi +\frac{5\pi }{6}$

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