Answer :


Ideas required to solve the problem:


The general solution of any trigonometric equation is given as –


• sin x = sin y, implies x = nπ + (– 1)ny, where n Z.


• cos x = cos y, implies x = 2nπ ± y, where n Z.


• tan x = tan y, implies x = nπ + y, where n Z.


Given,


cos x cos 2x cos 3x = �


4cos x cos 2x cos 3x – 1 = 0


{ 2 cos A cos B = cos (A + B) + cos (A – B)}


2(2cos x cos 3x)cos 2x – 1 = 0


2(cos 4x + cos 2x)cos2x – 1 = 0


2(2cos2 2x – 1 + cos 2x)cos 2x – 1 = 0 {using cos 2θ = 2cos2θ – 1 }


4cos3 2x – 2cos 2x + 2cos2 2x – 1 = 0


2cos2 2x (2cos 2x + 1) -1(2cos 2x + 1) = 0


(2cos2 2x – 1)(2 cos 2x + 1) = 0


either, 2cos 2x + 1 = 0 or (2cos2 2x – 1) = 0


cos 2x = -1/2 or cos 4x = 0 {using cos 2θ = 2cos2θ – 1}


cos 2x = cos (π - π/3) = cos 2π /3 or cos 4x = cos π/2


If cos x = cos y implies x = 2nπ ± y, where n Z.


In case of cos x = 0 we can give solution directly as cos x = 0 is true for x = odd multiple of π/2


Comparing obtained equation with standard equation, we have:


y = 2π / 3 or y = π/2


2x = 2mπ ± 2π/3 or 4x = (2n+1)π/2


where m,n ϵ Z ….ans


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