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) = 2

Now, let's use the identity sin^2(x) + cos^2(x) = 1 to substitute sin^2(x) with 1 - cos^2(x):
(1 - cos^2(x)) - 2√3sin(x)cos(x) + 3cos^2(x) = 2

Rearranging the terms, we get:
4cos^2(x) - 2√3sin(x)cos(x) - 1 = 0

Now, let's factor the quadratic equation:
(2cos(x) - √3sin(x))(2cos(x) + √3sin(x)) = 1

Since the product of two factors is equal to 1, either both factors are equal to 1 or one factor is equal to -1 and the other factor is equal to -1/1 = -1.

Case 1: 2cos(x) - √3sin(x) = 1
Rearranging the equation, we get:
2cos(x) = 1 + √3sin(x)
cos(x) = (1 + √3sin(x))/2

Case 2: 2cos(x) + √3sin(x) = 1
Rearranging the equation, we get:
2cos(x) = 1 - √3sin(x)
cos(x) = (1 - √3sin(x))/2

Now, we can solve each case separately.

Case 1: cos(x) = (1 + √3sin(x))/2
Squaring both sides of the equation, we get:
cos^2(x) = (1 + √3sin(x))^2/4
1 - sin^2(x) = (1 + 2√3sin(x) + 3sin^2(x))/4
4 - 4sin^2(x) = 1 + 2√3sin(x) + 3sin^2(x)
7sin^2(x) + 2√3sin(x) - 3 = 0

Now, we can solve this quadratic equation for sin(x) using the quadratic formula:
sin(x) = (-2√3 ± √(4√3^2 - 4(7)(-3)))/(2(7))
sin(x) = (-2√3 ± √(12 + 84))/(14)
sin(x) = (-2√3 ± √96)/(14)
sin(x) = (-√3 ± √24)/(7)

Case 2: cos(x) = (1 - √3sin(x))/2
Squaring both sides of the equation, we get:
cos^2(x) = (1 - √3sin(x))^2/4
1 - sin^2(x) = (1 - 2√3sin(x) + 3sin^2(x))/4
4 - 4sin^2(x) = 1 - 2√3sin(x) + 3sin^2(x)
3sin^2(x) + 2√3sin(x) - 3 = 0

Now, we can solve this quadratic equation for sin(x) using the quadratic formula:
sin(x) = (-2√3 ± √(4√3^2 - 4(3)(-3)))/(2(3))
sin(x) = (-2√3 ± √(12 + 36))/(6)
sin(x) = (-2√3 ± √48)/(6)
sin(x) = (-√3 ± √16)/(3)

Therefore, the solutions for the equation sin(x) - √3cos(x) = √2 are:
sin(x) = (-√3 ± √24)/(7)
sin(x) = (-√3 ± √16)/(3)