$1 + cos 2 x + cos 4 x + cos 6 x - 4 cos x cos 2 x cos 3 x =$

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Solution

The correct option is D 0
consider $1 + cos 2 x + cos 4 x + cos 6 x - 4 cos x cos 2 x cos 3 x$

$= (1 + cos 2 x) + (cos 4 x + cos 6 x) - 4 cos x cos 2 x cos 3 x$

$= 2 {cos}^{2} x + 2 cos (\frac{4 x + 6 x}{2}) cos (\frac{4 x - 6 x}{2}) - 4 cos x cos 2 x cos 3 x$

$= 2 {cos}^{2} x + 2 cos (\frac{10 x}{2}) cos (\frac{- 2 x}{2}) - 4 cos x cos 2 x cos 3 x$

$= 2 {cos}^{2} x + 2 cos 5 x cos x - 4 cos x cos 2 x cos 3 x$

$= 2 cos x (cos x + cos 5 x) - 4 cos x cos 2 x cos 3 x$

$= 2 cos x (2 cos (\frac{x + 5 x}{2}) cos (\frac{x - 5 x}{2})) - 4 cos x cos 2 x cos 3 x$

$= 4 cos x (cos (\frac{6 x}{2}) cos (\frac{- 4 x}{2})) - 4 cos x cos 2 x cos 3 x$

$= 4 cos x cos 2 x cos 3 x - 4 cos x cos 2 x cos 3 x = 0$

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