To solve this trigonometric equation, we can use the difference of angles formula for sine and cosine:

sin(a - b) = sin(a)cos(b) - cos(a)sin(b)

Substitute a = 7x and b = 5x:

sin(7x - 5x) = sin(7x)cos(5x) - cos(7x)sin(5x)
sin(2x) = sin(7x)cos(5x) - cos(7x)sin(5x)

Since sin(2x) = 2sin(x)cos(x), we have:

2sin(x)cos(x) = sin(7x)cos(5x) - cos(7x)sin(5x)

Now, expand the right side using the product-to-sum identities:

sin(7x)cos(5x) - cos(7x)sin(5x) = sin(7x)cos(5x) - sin(7x)sin(5x)
= sin(7x)(cos(5x) - sin(5x))

Now, we can express cos(5x) - sin(5x) as sin(5x + π/4) using the sum-to-product formula for sine:

sin(a + b) = sin(a)cos(b) + cos(a)sin(b)

Let a = 5x and b = π/4:

sin(5x + π/4) = sin(5x)cos(π/4) + cos(5x)sin(π/4)
= sin(5x)(1/√2) + cos(5x)(1/√2)
= [sin(5x) + cos(5x)]/√2

So, our equation becomes:

2sin(x)cos(x) = sin(7x)(sin(5x) + cos(5x))/√2

simplify:

2sin(x)cos(x) = [sin(7x)sin(5x) + sin(7x)cos(5x)]/√2
2sin(x)cos(x) = sin(7x + 5x)/√2
2sin(x)cos(x) = sin(12x)/√2
2√2sin(x)cos(x) = sin(12x)

Now we have a simpler equation to solve:

2√2sin(x)cos(x) = sin(12x)

sin(12x) = 2√2sin(x)cos(x)

Since sin(12x) = 2sin(6x)cos(6x), we have:

2sin(6x)cos(6x) = 2√2sin(x)cos(x)

simplify:

sin(6x)cos(6x) = √2sin(x)cos(x)

sin(2(6x)) = √2sin(2x)

sin(12x) = √2sin(2x)

Since sin(12x) = sin(2x + 10x), we must have:

2x + 10x = 12x
12x = 12x

Therefore, the equation is true for all values of x.