Question

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

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

Answer 1

The correct option is C $\frac{\pi }{3}(6k+1),k\in Z$

$\cos 3x+\sin (2x-\frac{7\pi }{6})=-2$
$\Rightarrow 1+\cos 3x+1+\sin (2x-\frac{7\pi }{6})=0$
$\Rightarrow (1+\cos 3x)+1-\cos (2x-\frac{2\pi }{3})=0$
$\Rightarrow 2{\cos }^2\frac{3x}{2}+2{\sin }^2(x-\frac{\pi }{3})=0$
$\Rightarrow \cos \frac{3x}{2}=0$ and $\sin (x-\frac{\pi }{3})=0$
$\Rightarrow \frac{3x}{2}=\frac{\pi }{2},\frac{3\pi }{2},\cdots$
and $x-\frac{\pi }{3}=0,\pi ,2\pi ,\cdots$

Therefore, the general solution is $x=2k\pi +\frac{\pi }{3}=\frac{\pi }{3}(6k+1),k\in Z$