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 = 2 sin \((\frac{A+B}2)\) cos \((\frac{A-B}2)\)

⇒ sin 2x + 2 sin \((\frac{3x+x}2)\) cos \((\frac{3x-x}2)\) = 0

⇒ 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 = - 1/2 = cos (π-\(\frac{π}3\)) = cos \(\frac{2π}3\)

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 ± \(\frac{2π}3\)

∴ x = \(\frac{nπ}2\) or x = 2mπ  ± \(\frac{2π}3\)where m,n ϵ Z ..ans