To solve this equation, we can first simplify it by using the trigonometric identity sin^2x + cos^2x = 1.

So, the equation becomes:

3sin^2x + 2√3sinxcosx + cos^2x = 0
= 3sin^2x + 2√3sinxcosx + 1 = 0
= (sinx + √3cosx)^2 = 0

Since the square of any real number is always non-negative, the only way that the expression can be equal to zero is if the square itself is zero.

Therefore, sinx + √3cosx = 0

Now we can solve for x in terms of trigonometric functions.

sinx + √3cosx = 0
sinx = -√3cosx
tanx = -√3
x = arctan(-√3) + nπ, where n is an integer

Therefore, the solution to the given equation is x = arctan(-√3) + nπ, where n is an integer.