What is the normal form of the line $x + \sqrt{3}y - 4 = 0$?
- A$x \cos(\pi/3) - y \sin(\pi/3) = 2$
- B$x \cos(\pi/6) - y \sin(\pi/6) = 2$
- C$x \cos(\pi/3) + y \sin(\pi/3) = 2$
- D$x \cos(\pi/6) + y \sin(\pi/6) = 2$
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