To solve the equation sin(x) - √3cos(x) = 2, we can use the trigonometric identity sin^2(x) + cos^2(x) = 1.

First, let's square both sides of the equation:
(sin(x) - √3cos(x))^2 = 2^2
sin^2(x) - 2√3sin(x)cos(x) + 3cos^2(x) = 4

Using the trigonometric identity sin^2(x) + cos^2(x) = 1, we can substitute 1 - cos^2(x) for sin^2(x) in the equation:
(1 - cos^2(x)) - 2√3sin(x)cos(x) + 3cos^2(x) = 4

Rearranging the terms, we get:
4cos^2(x) - 2√3sin(x)cos(x) - cos^2(x) = 3

Combining like terms, we have:
3cos^2(x) - 2√3sin(x)cos(x) - 3 = 0

Now, let's substitute u = cos(x):
3u^2 - 2√3sin(x)u - 3 = 0

This is a quadratic equation in terms of u. We can solve it using the quadratic formula:
u = (-b ± √(b^2 - 4ac)) / (2a)

Plugging in the values, we have:
u = (-(-2√3sin(x)) ± √((-2√3sin(x))^2 - 4(3)(-3))) / (2(3))
u = (2√3sin(x) ± √(12sin^2(x) + 36)) / 6
u = (√3sin(x) ± √(sin^2(x) + 3)) / 3

Since u = cos(x), we can substitute back:
cos(x) = (√3sin(x) ± √(sin^2(x) + 3)) / 3

Now, we can solve for sin(x) using the identity sin^2(x) + cos^2(x) = 1:
sin^2(x) = 1 - cos^2(x)
sin^2(x) = 1 - ((√3sin(x) ± √(sin^2(x) + 3)) / 3)^2

Expanding and simplifying, we get:
sin^2(x) = 1 - (3sin^2(x) ± 2√3sin(x)√(sin^2(x) + 3) + (sin^2(x) + 3)) / 9
9sin^2(x) = 9 - 3sin^2(x) ± 2√3sin(x)√(sin^2(x) + 3) - (sin^2(x) + 3)
12sin^2(x) ± 2√3sin(x)√(sin^2(x) + 3) = 6

Dividing both sides by 2, we have:
6sin^2(x) ± √3sin(x)√(sin^2(x) + 3) = 3

Squaring both sides, we get:
36sin^4(x) ± 12√3sin^3(x)√(sin^2(x) + 3) + 18sin^2(x) = 9

Rearranging the terms, we have:
36sin^4(x) ± 12√3sin^3(x)√(sin^2(x) + 3) + 18sin^2(x) - 9 = 0

This is a quartic equation in terms of sin(x). Unfortunately, there is no simple algebraic solution to this equation. It can be solved numerically or graphically to find the approximate values of sin(x) that satisfy the equation.