To solve the equation sin(3x) + sin(7x) = 0, we can use the sum-to-product trigonometric identity:

sin(a) + sin(b) = 2sin((a+b)/2)cos((a-b)/2)

We have:

sin(3x) + sin(7x) = 2sin((3x+7x)/2)cos((3x-7x)/2)
sin(3x) + sin(7x) = 2sin(5x)cos(-2x)
sin(3x) + sin(7x) = 2sin(5x)cos(2x)

Now, we want to find the values of x that satisfy sin(3x) + sin(7x) = 0:

2sin(5x)cos(2x) = 0

This equation will be satisfied when either sin(5x) or cos(2x) is equal to 0.

First, let's look at sin(5x) = 0:

sin(5x) = 0
5x = nπ, where n is an integer
x = nπ/5, where n is an integer

Now, let's look at cos(2x) = 0:

cos(2x) = 0
2x = (2n+1)π/2, where n is an integer
x = (2n+1)π/4, where n is an integer

Therefore, the solutions to the equation sin(3x) + sin(7x) = 0 are x = nπ/5 and x = (2n+1)π/4, where n is an integer.