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 2x - sin 4x + sin 6x = 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.

we have, sin 2x - sin 4x + sin 6x = 0

∴ we will use sin 6x and sin 2x for transformation as after transformation it will give sin 4x term which can be taken common.

{∵ sin A + sin B =

⇒ -sin 4x + 2 sin

⇒ 2sin 4x cos 2x – sin 4x = 0

⇒ sin 4x ( 2cos 2x – 1) = 0

∴ either, sin 4x = 0 or 2cos 2x – 1 = 0

⇒ sin 4x = sin 0 or cos 2x = � = cos π/3

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

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

Comparing obtained equation with standard equation, we have:

4x = nπ or 2x = 2mπ ± π/3

∴ where m,n ϵ Z ..ans