Q. 4 B5.0( 1 Vote )

Solve the following equations :

cos x + cos 3x – cos 2x = 0

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 = 0


To solve the equation we need to change its form so that we can equate the t-ratios individually.


For this we will be applying transformation formulae. While applying the


Transformation formula we need to select the terms wisely which we want


to transform.


As, cos x - cos 2x + cos 3x = 0


we will use cos x and cos 2x for transformation as after transformation it will give cos 2x term which can be taken common.


cos x - cos 2x + cos 3x = 0


-cos 2x + (cos x + cos 3x) = 0


{ cos A + cos B = 2


-cos 2x + 2 cos


-cos 2x + 2cos 2x cos x = 0


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


cos 2x = 0 or 1 + 2cos x = 0


cos 2x = cos π/2 or cos x = 1/2


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


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


From above expression and on comparison with standard equation we have:


y = π/2 or y = π/3


2x = 2nπ ± π/2 or x = 2mπ ± π/3


where m, n ϵ Z


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