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 = sinx
∴ sin x + sin 5x – sin 3x = 0
∴ we will use sin x and cos 5x for transformation as after transformation it will give sin 3x term which can be taken common.
{∵ sin A + sin B = 2 sin \((\frac{A+B}2)\) cos \((\frac{A-B}2)\)]
⇒ - sin 3x + 2 sin \((\frac{5x+x}2)\) cos \(\frac{5x-x}2\) = 0
⇒ 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 = \(\frac{1}{2}\) = cos \(\frac{π}3\)
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 ± \(\frac{π}3\)
∴ x = \(\frac{nπ}3\) or x = mπ ± \(\frac{π}6\)where m,n ϵ Z ..ans
