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


we will use sin x and sin 3x together in 1 group for transformation and sin 4x and sin 2x common in other group as after transformation both will give cos x term which can be taken common.


{ sin A + sin B =


(sin x + sin 3x )+( sin 2x + sin 4x) = 0


2 sin + 2 sin


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


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


Again using transformation formula, we have:


2 cos x 2



either, cos x = 0 or or


In case of cos x = 0 we can give solution directly as cos x = 0 is true for x = odd multiple of π/2


In case of sin x = 0 we can give solution directly as sin x = 0 is true for x = integral multiple of π



where n,p,m ϵ Z


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