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,


3 – 2 cos x – 4 sin x – cos 2x + sin 2x = 0


As, cos 2x = 1 – 2sin2 x and sin 2x = 2sin x cos x


3 – 2cos x – 4sin x – (1 – 2sin2 x) + 2sin x cos x = 0


2sin2 x – 4sin x + 2 – 2cos x + 2sin x cos x = 0


2(sin2 x – 2sin x + 1) + 2cos x(sin x – 1) = 0


2(sin x – 1)2 + 2cos x(sin x – 1) = 0


(sin x – 1)(2cos x + 2sin x – 2) = 0


sin x = 1 or sin x + cos x = 1


When, sin x = 1


We have,


sin x = sin π/2


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



When, sin x + cos x = 1


{ dividing by √2 both sides}


{ }


{ cos A cos B + sin A sin B = cos (A - B)}


If cos x = cos y, implies x = 2mπ ± y, where m Z





Hence,



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