Question

The equation $8\cos x.\cos 2x.\cos 4x=\frac{\sin 6x}{\sin x}$ has solution, if

• A. $\sin x=0$
• B. $\cos 7x=0$
• C. $\sin 7x=0$
• D. $\sin 8x=0$

Answer 1

Answer: (B)

$\cos 7x=0$

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

We know $\sin 2x=2\sin x.\cos x$

Using this, we have
$4\left(2\cos x\sin x\right)\cos 2x\cos 4x=\sin 6x$
$2\left(2\sin 2x\cos 2x\right)\cos 4x=\sin 6x$

$2\left(\sin 4x\right)\cos 4x=\sin 6x$
$\sin 8x=\sin 6x$
$2\cos 7x\sin x=0$
But $\sin x$ is non zero, so $\cos 7x=0$.