Question

4 cos x cos 2x. cos3x = 1. solve it. ​

Answer 1

Answer:

$x = (nπ ± π/3) \:$

Step-by-step explanation:

Given:-

4 cos x * cos 2x * cos 3x = 1

=> (2 cos x * cos 3x) * (2 cos 2x) = 1

=> {cos (3x+ x) + cos (3x - x)} * (2 cos 2x) = 1

=> (cos 4x + cos 2x) * (2 cos 2x) = 1

=> 2 cos 4x * cos 2x + 2 cos² 2x = 1

=> 2 cos 4x * cos 2x + (2 cos² 2x - 1) = 0

=> 2 cos 4x * cos 2x + cos 4x = 0

=> cos 4x (2 cos 2x + 1) = 0

=> (i) cos 4x = 0, or (ii) 2 cos 2x + 1 = 0

(i) cos 4x = 0

=> cos 4x = cos (2nπ ± π/2)

=> 4x = (2nπ ± π/2) = (4n ± 1)π/2

=> x = (4n ± 1)π/8

(ii) 2 cos 2x + 1 = 0

=> 2 cos2x = - 1

=> cos 2x = - 1/2 = cos (2nπ ± 2π/3)

=> 2x = (2nπ ± 2π/3)

=> x = (nπ ± π/3)

where n = 0, 1, 2, 3 .......

Hope it helps you.

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