Question

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

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

Answer 1

Answer: (A)

We have $\cos 3x+\sin \left(2x-\frac{7\pi }{6}\right)=-2$
$\Rightarrow 1+\cos 3x+1+\sin \left(2x-\frac{7\pi }{6}\right)=0$
$\Rightarrow (1+\cos 3x)+1-\cos \left(2x-\frac{2\pi }{3}\right)=0$
$\Rightarrow 2{\cos }^2\frac{3x}{2}+2{\sin }^2\left(x-\frac{\pi }{3}\right)=0$
$\Rightarrow \cos \frac{3x}{2}=0$ and $\sin \left(x-\frac{\pi }{3}\right)=0$
$\Rightarrow \frac{3x}{2}=\frac{\pi }{2},\frac{3\pi }{2},\dots$ and $x-\frac{\pi }{3}=0,\pi ,2\pi ,\dots$
$\Rightarrow x=\frac{\pi }{3}$
Therefore, the general solution to $\cos \frac{3x}{2}=0$
and $\sin \left(x-\frac{\pi }{3}\right)=0$ is
$x=2k\pi +\frac{\pi }{3}=\frac{\pi }{3}(6k+1)$, where $k\in Z$