Question

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

• A. $x=\left(\frac{n\pi }{7}\right)+\left(\frac{\pi }{21}\right),\forall nϵz$
• B. $x=\left(\frac{2\pi }{7}\right)+\left(\frac{\pi }{14}\right),\forall nϵz$
• C. $x=\left(\frac{n\pi }{7}\right)+\left(\frac{\pi }{14}\right),\forall nϵz$
• D. None of these

Answer 1

The correct option is D None of these

$8\cos x\cos 2x\cos 4x=\frac{\sin 6x}{\sin x}$

$4\sin 2x\cos 2x\cos 4x=\sin 6x$

$2\sin 4x\cos 4x=\sin 6x$

$\sin 8x-\sin 6x=0$

$2\cos \frac{7x}{2}\sin x=0$

$\Rightarrow \cot \frac{7x}{2}\sin x=0$

The general solutions are

$\frac{7x}{2}=n\pi +\frac{\pi }{2},n\in I$ or $x=m\pi ,m\in I$