Question

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

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

Answer 1

Answer: (A)

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