To prove the given equation, we can use the product-to-sum trigonometric identity:

sinAcosB = (1/2)[sin(A + B) + sin(A - B)]

Let A = 3x and B = x:

sin(3x)cos(x) = (1/2)[sin(3x + x) + sin(3x - x)]
sin(3x)cos(x) = (1/2)[sin(4x) + sin(2x)]

Applying the identity for sin(2x):

sin(3x)cos(x) = (1/2)[sin(4x) + 2sin(x)cos(x)]

Now, we can substitute sin(4x) with 2sin(2x)cos(2x):

sin(3x)cos(x) = (1/2)[2sin(2x)cos(2x) + 2sin(x)cos(x)]

Apply the double angle formula for sin(2x) and cos(2x):

sin(3x)cos(x) = sin(x)(2cos^2(x) - 1) + 2sin(x)cos^2(x)

sin(3x)cos(x) = 2sin(x)cos^2(x) - sin(x) + 2sin(x)cos^2(x)

sin(3x)cos(x) = 4sin(x)cos^2(x) - sin(x)

sin(3x)cos(x) - sin(x)cos(3x) = sin(x)

Therefore, sin(3x)cos(x) - sin(x)cos(3x) = 1.