If 1 + cos 2x + cos 4x + cos 6x = k cos x cos 2x cos 3x,

Solution:

Given 1 + cos 2x + cos 4x + cos 6x = k cos x cos 2x cos 3x

Consider,

L.H.S $1+\cos 2 x+\cos 4 x+\cos 6 x$$=(1+\cos 2 x)+(\cos 4 x+\cos 6 x)$

$=2 \cos ^{2} x+2 \cos \left(\frac{4 x+6 x}{2}\right) \cos \left(\frac{4 x-6 x}{2}\right)$

using identities : $-1+\cos 2 \theta=2 \cos ^{2} \theta$ and $\cos a+\cos b=2 \cos \left(\frac{a+b}{2}\right) \cos \left(\frac{a-b}{2}\right)$

$=2 \cos ^{2} x+2 \cos 5 x \cos (-x)$

$=2 \cos ^{2} x+2 \cos 5 x \cos x$           $(\because \cos (-x)=\cos x)$

$=2 \cos x(\cos x+\cos 5 x)$

$=2 \cos x\left[2 \cos \left(\frac{x+5 x}{2}\right) \cos \left(\frac{x-5 x}{2}\right)\right]$

$=2 \cos x[2 \cos 3 x \cos (-2 x)]$

$=2 \cos x[2 \cos 3 x \cos 2 x]$          $(\because \cos (-\theta)=\cos \theta)$

$=4 \cos x \cos 2 x \cos 3 x$

Since L.H.S = R. H. S is given,

$\Rightarrow 4 \cos x \cos 2 x \cos 3 x=k \cos x \cos 2 x \cos 3 x$

$\Rightarrow k=4$ is the answer