Solve the following equations :

sin x + sin 2x + sin 3x = 0


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 + sin 2x + sin 3x = 0


To solve the equation we need to change its form so that we can equate the t-ratios individually.


For this we will be applying transformation formulae. While applying the


Transformation formula we need to select the terms wisely which we want


to transform.


As, sin x + sin 2x + sin 3x = 0


we will use sin x and sin 3x for transformation as after transformation it will give sin 2x term which can be taken common.


{ sin A + sin B =


sin 2x + 2 sin


2sin 2x cos x + sin 2x = 0


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


either, sin 2x = 0 or 2cos x + 1 = 0


sin 2x = sin 0 or cos x = - � = cos (π-π/3) = cos 2π/3


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


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


Comparing obtained equation with standard equation, we have:


2x = nπ or x = 2mπ ± 2π/3


where m,n ϵ Z ..ans


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