I suggest you to transform the difference of like trigonometric functions into a product such that:

sin alpha - sin beta = 2 cos ((alpha+beta)/2)*(sin(alpha-beta)/2)

sin 4x - sin2x = 2 cos ((4x+2x)/2)*(sin(4x-2x)/2)

sin 4x - sin2x = 2 cos (3x)*sin(-x)

sin 4x - sin2x = - 2 cos (3x)*sin(x)

Replacing the difference in equation...

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I suggest you to transform the difference of like trigonometric functions into a product such that:

sin alpha - sin beta = 2 cos ((alpha+beta)/2)*(sin(alpha-beta)/2)

sin 4x - sin2x = 2 cos ((4x+2x)/2)*(sin(4x-2x)/2)

sin 4x - sin2x = 2 cos (3x)*sin(-x)

sin 4x - sin2x = - 2 cos (3x)*sin(x)

Replacing the difference in equation sin 4x - sin2x = 0 yields:

- 2 cos (3x)*sin(x) = 0 => cos 3x = 0 and sin x = 0

cos 3x = 0 => 3x = pi/2 + 2n*pi => x = pi/6 + 2n*(pi/3)

sin x = 0 => x = 0

The solutions to trigonometric equation are x=0; x=pi/6 + 2n*(pi/3).