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,


4 sin x cos x + 2 sin x + 2 cos x + 1 = 0


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


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


cos x = -1/2 or sin x = -1/2


cos x = cos (π - π/3) or sin x = sin (- π/6)


cos x = cos 2π/3 or sin x = sin (-π/6)


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


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


x = 2nπ ± 2π/3 or x = mπ + (-1)m (-π/6)


Hence,



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