To solve the equation sqrt(3)sin^2x + sinx*cosx = 0, we can first factor out sinx from the equation:

sinx(sqrt(3)sinx + cosx) = 0

Now we have two possible solutions:

For sinx = 0, the solution is x = nπ, where n is an integer.

For the second equation sqrt(3)sinx + cosx = 0, we can square both sides to get rid of the square root:

(√3sinx)^2 + 2√3sinxcosx + (cosx)^2 = 0
3sin^2x + 2√3sinxcosx + cos^2x = 0

Now we have a quadratic equation in terms of sinx and cosx. Let's substitute sinx = a and cosx = b:

3a^2 + 2√3ab + b^2 = 0
(3a + b)(a + √3b) = 0

This equation has two possible solutions:

Plugging back sinx = a and cosx = b, we get:

3sinx + cosx = 0
3sinx = -cosx
tanx = -1/3
x = arctan(-1/3)

sinx + √3cosx = 0
sinx = -√3cosx
tanx = -√3
x = arctan(-√3)

Therefore, the solutions to the equation sqrt(3)sin^2x + sinx*cosx = 0 are x = nπ, x = arctan(-1/3), and x = arctan(-√3) where n is an integer.