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,


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


(sin x + sin 3x) – 3sin 2x – (cos x + cos 3x) + 3 cox 2x = 0


sin A + sin B =



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


sin 2x ( 2cos x – 3) - cos 2x (2cos x – 3) = 0


(2cos x – 3)(sin 2x – cos 2x) = 0


cos x = 3/2 = 1.5 (not accepted as cos x lies between – 1 and 1)


Or sin 2x = cos 2x


tan 2x = 1 = tan π/4


If tan x = tan y, implies x = nπ + y, where n Z.


2x = nπ + π/4



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