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 5x = sin 3x

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 5x = sin 3x

∴ sin x + sin 5x – sin 3x = 0

∴ we will use sin x and sin 5x for transformation as after transformation it will give sin 3x term which can be taken common.

{∵ sin A + sin B =

⇒ -sin 3x + 2 sin

⇒ 2sin 3x cos 2x – sin 3x = 0

⇒ sin 3x ( 2cos 2x – 1) = 0

∴ either, sin 3x = 0 or 2cos 2x – 1 = 0

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

3x = nπ or 2x = 2mπ ± π/3

∴ where m,n ϵ Z ..ans