solve:-
$1 + cos 2 x + cos 4 x + cos 6 x =$

2 cos x cos 2 x c o s 3 x

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2 cos x sin 2 x c o s 3 x

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4 cos x cos 2 x c o s 3 x

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4 cos x sin 2 x sin 3 x

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Solution

The correct option is C $4 cos x cos 2 x c o s 3 x$
$1 + cos 2 x + cos 4 x + cos 6 x = (1 + cos 2 x) + (cos 4 x + cos 6 x)$
$= 2 {cos}^{2} x + 2 cos 5 x cos x$ $(∵ cos A + cos B = 2 cos \frac{A + B}{2} cos \frac{A - B}{2}, cos 2 A = 2 {cos}^{2} A - 1)$
$= 2 cos x (cos x + cos 5 x)$
$= 2 cos x (2 cos 2 x cos 3 x)$
$= 4 cos x cos 2 x cos 3 x$

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