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)n y, 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,


2 cos2 x – 5 cos x + 2 =0


As the equation is of 2nd degree, so we need to solve a quadratic equation.


First we will substitute trigonometric ratio with some variable k and we will solve for k


Let, cos x = k


2k2 – 5k + 2 = 0


2k2 – 4k – k +2 = 0


2k(k – 2) -1(k -2) = 0


(k – 2)(2k - 1) = 0


k = 2 or k = �


cos x = 2 {which is not possible} or cos x = � (acceptable)


cos x = �


cos x = cos 60° = cos π/3


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


On comparing our equation with standard form, we have


y = π/3


where n ϵ Z ..ans


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