To solve this equation, we can rewrite it in terms of the double angle formula:

sin^2x + 6sinxcosx + 8cos^2x = 0
(sin^2x + 2sinxcosx + cos^2x) + 4sinxcosx + 4cos^2x = 0
(sin(x + y))^2 + 4sin(2x) + 4cos^2x = 0
(sin(x + y))^2 + 4sin(2x) + 4(1 - sin^2x) = 0

Letting u = sin(x + y), the equation becomes:

u^2 + 4(2u) + 4(1 - u^2) = 0
u^2 + 8u + 4 - 4u^2 = 0
-3u^2 + 8u + 4 = 0

Solving this quadratic equation gives us:

u = (-b ± √(b^2 - 4ac)) / 2a
u = (-8 ± √(8^2 - 4(-3)(4))) / 2(-3)
u = (-8 ± √(64 + 48)) / -6
u = (-8 ± √112) / -6
u = (-8 ± 4√7) / -6

Taking the positive value (since u = sin(x + y)), we get:

u = (-8 + 4√7) / -6

Now, we can solve for sin(x + y):

sin(x + y) = (-8 + 4√7) / -6

Therefore, the solution to the equation sin^2x + 6sinxcosx + 8cos^2x = 0 is sin(x + y) = (-8 + 4√7) / -6.