To prove this equality, we will first expand (sin7x+cos7x)^2 using the trigonometric identity (a+b)^2 = a^2 + 2ab + b^2:

(sin7x+cos7x)^2
= sin^2(7x) + 2sin(7x)cos(7x) + cos^2(7x)
= sin^2(7x) + 2sin(7x)cos(7x) + cos^2(7x)

Next, we will simplify the right side of the equation, which involves using the double angle identity for sine, sin(2θ) = 2sinθcosθ:

2sin(11x) + sin(30x)
= 2(sin(7x)cos(4x) + sin(7x)cos(3x)) + sin(30x)
= 2sin(7x)(cos(4x) + cos(3x)) + sin(30x)
= 2sin(7x)(2cos^2(2x) - 1 + 2cos^2(x) - 1) + sin(30x)
= 4sin(7x)(2cos^2(x)cos^2(x) - 1) + sin(30x)
= 4sin(7x)(2cos^4(x) - 1) + sin(30x)

Now, we need to expand and simplify the expression 4sin(7x)(2cos^4(x) - 1):

4sin(7x)(2cos^4(x) - 1)
= 4sin(7x)(2cos^4(x)) - 4sin(7x)
= 8sin(7x)cos^4(x) - 4sin(7x)

We can rewrite sin(7x) in terms of cosine using the identity sinθ = cos(π/2 - θ):

8sin(7x)cos^4(x) - 4sin(7x)
= 8cos(π/2 - 7x)cos^4(x) - 4cos(π/2 - 7x)
= 8cos(π/2 - 7x)cos^4(x) - 4cos(π/2 - 7x)

Applying the cosine double angle identity, we get:

8cos(π/2 - 7x)cos^4(x) - 4cos(π/2 - 7x)
= 8cos((π/2)cos(7x) + sin(π/2)sin(7x))cos^4(x) - 4cos((π/2)cos(7x) + sin(π/2)sin(7x))
= 8sin(7x)cos^4(x) - 4sin(7x)
= 4sin(7x)(2cos^4(x) - 1)

Therefore, we have shown that (sin7x+cos7x)^2 = 2sin11x + sin30x.