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

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Solution

Given 1 + cos 2x + cos 4x + cos 6x = k cos x cos 2x cos 3x
Consider,
$L . H . S 1 + \cos 2 x + \cos 4 x + \cos 6 x = (1 + \cos 2 x) + (\cos 4 x + \cos 6 x) = 2 \cos^{2} x + 2 \cos (\frac{4 x + 6 x}{2}) \cos (\frac{4 x - 6 x}{2}) using identities : - 1 + \cos 2 θ = 2 \cos^{2} θ and \cos a + \cos b = 2 \cos (\frac{a + b}{2}) \cos (\frac{a - b}{2}) = 2 \cos^{2} x + 2 \cos 5 x \cos (- x) = 2 \cos^{2} x + 2 \cos 5 x \cos x (∵ \cos (- x) = \cos x) = 2 \cos x (\cos x + \cos 5 x)$

$= 2 \cos x [2 \cos (\frac{x + 5 x}{2}) \cos (\frac{x - 5 x}{2})] = 2 \cos x [2 \cos 3 x \cos (- 2 x)] = 2 \cos x [2 \cos 3 x \cos 2 x] (∵ \cos (- θ) = \cos θ) = 4 \cos x \cos 2 x \cos 3 x Since L . H . S = R . H . S is given, \Rightarrow 4 \cos x \cos 2 x \cos 3 x = k \cos x \cos 2 x \cos 3 x \Rightarrow k = 4 is the answer$

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