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,


2 sin2 x = 3 cos x , 0 ≤ x ≤ 2π


2 ( 1 – cos2 x) = 3 cos x


2 cos2 x + 3cos x – 2 = 0


2 cos2 x + 4 cos x – cos x – 2 = 0


2 cos x(cos x + 2) – 1(cos x + 2) = 0


(2cos x – 1)(cos x + 2) = 0


cos x = �


or cos x = -2 { as cos x lies between -1 and 1 so this value is rejected }


cos x = � = cos π/3


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


x = 2nπ ± π/3


But, 0 ≤ x ≤ 2π


x = π/3 and x = 2π - π/3 = 5π/3 ….ans


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