Question

The general solution of $8\cos x\cdot \cos 2x\cdot \cos 4x=\frac{\sin 6x}{\sin x}$ is

• A. $\frac{(2n+1)\pi }{14},n\in Z$
• B. $\left\{n\pi \right\}\cup \left\{\frac{(2n+1)\pi }{14}\right\},n\in Z$
• C. $\frac{(2n+1)\pi }{7},n\in Z$
• D. $\frac{n\pi }{7},n\in Z$

Answer 1

The correct option is A $\frac{(2n+1)\pi }{14},n\in Z$

$8\cos x\cdot \cos 2x\cdot \cos 4x=\frac{\sin 6x}{\sin x}$
$\sin x\ne 0$
Now,
$\Rightarrow 8\cos x\cdot \sin x\cdot \cos 2x\cdot \cos 4x-\sin 6x=0\Rightarrow 4\sin 2x\cdot \cos 2x\cdot \cos 4x-\sin 6x=0\Rightarrow 2\sin 4x\cdot \cos 4x-\sin 6x=0\Rightarrow \sin 8x-\sin 6x=0\Rightarrow 2\cos 7x\sin x=0\Rightarrow \cos 7x=0(\because \sin x\ne 0)\Rightarrow 7x=\frac{(2n+1)\pi }{2},n\in Z\therefore x=\frac{(2n+1)\pi }{14},n\in Z$