If $1 + cos 2 x + cos 4 x + cos 6 x = y cos x cos 2 x cos 3 x$ , then find the value of $y$ .

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Solution

$1 + cos 2 x + cos 4 x + cos 6 x =$
$(1 + cos 6 x) + (cos 2 x + cos 4 x)$

$(2 cos 3 x)^{2} + 2 cos 3 x cos x$

$2 cos 3 x (cos 3 x + cos x)$

$2 cos 3 x .2 cos 2 x . cos x$

$= 4 cos x . cos 2 x . cos 3 x$

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