Step 1 :First, we rewrite the equation \(\sin \frac{x}{3}-\cos \frac{x}{3}=0\) as \(\sin \frac{x}{3}=\cos \frac{x}{3}\).
Step 2 :We know that \(\sin \theta = \cos \theta\) when \(\theta = \frac{\pi}{4} + n\pi\), where n is an integer.
Step 3 :So, we can write \(\frac{x}{3} = \frac{\pi}{4} + n\pi\).
Step 4 :Solving for x, we get \(x = 3\left(\frac{\pi}{4} + n\pi\right)\).
Step 5 :We are looking for solutions in the interval [0, 2\pi), so we substitute n = 0, 1, 2, 3, 4, 5, 6, 7 into the equation and check which values of x fall within this interval.
Step 6 :For n = 0, we get \(x = 3\left(\frac{\pi}{4}\right) = \frac{3\pi}{4}\), which is in the interval [0, 2\pi).
Step 7 :For n = 1, we get \(x = 3\left(\frac{\pi}{4} + \pi\right) = \frac{7\pi}{4}\), which is in the interval [0, 2\pi).
Step 8 :For n = 2, we get \(x = 3\left(\frac{\pi}{4} + 2\pi\right) = \frac{11\pi}{4}\), which is not in the interval [0, 2\pi).
Step 9 :For n = 3, we get \(x = 3\left(\frac{\pi}{4} + 3\pi\right) = \frac{15\pi}{4}\), which is not in the interval [0, 2\pi).
Step 10 :For n = 4, we get \(x = 3\left(\frac{\pi}{4} + 4\pi\right) = \frac{19\pi}{4}\), which is not in the interval [0, 2\pi).
Step 11 :For n = 5, we get \(x = 3\left(\frac{\pi}{4} + 5\pi\right) = \frac{23\pi}{4}\), which is not in the interval [0, 2\pi).
Step 12 :For n = 6, we get \(x = 3\left(\frac{\pi}{4} + 6\pi\right) = \frac{27\pi}{4}\), which is not in the interval [0, 2\pi).
Step 13 :For n = 7, we get \(x = 3\left(\frac{\pi}{4} + 7\pi\right) = \frac{31\pi}{4}\), which is not in the interval [0, 2\pi).
Step 14 :So, the solutions to the equation \(\sin \frac{x}{3}-\cos \frac{x}{3}=0\) in the interval [0, 2\pi) are \(\boxed{\frac{3\pi}{4}, \frac{7\pi}{4}}\).