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

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Solution

The correct option is D $0$
$\Rightarrow (1 + cos 2 x) + (cos 6 x + cos 4 x) - 4 cos x cos 2 x cos 3 x$

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

$= 2 cos x [cos 5 x + cos x] - 4 cos x cos 2 x cos 3 x$

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

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

$= 0$

Hence, option $^{'} D^{'}$ is correct.

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