To solve the equation $\cot \left( \frac{3x}{3} \right) = 1$ on the interval $[0, 2\pi)$, let's proceed as follows:

1. Rewrite the Equation: The equation $\cot \left( \frac{3x}{3} \right) = 1$ simplifies to $\cot(3x) = 1$.

2. Solve for $3x$ Using Cotangent Values: The cotangent function equals 1 at specific angles within a period. We know: $\cot \left( \frac{\pi}{4} \right) = 1$ This means: $3x = \frac{\pi}{4} + n\pi \quad \text{for integer } n$

3. Solve for $x$: Divide both sides by 3 to isolate $x$: $x = \frac{\pi}{12} + \frac{n\pi}{3}$

4. Find All Solutions in $[0, 2\pi)$: Now we calculate specific values of $x$ by substituting integer values of $n$ to ensure $x$ is within $[0, 2\pi)$.

   • For $n = 0$: $x = \frac{\pi}{12}$
   • For $n = 1$: $x = \frac{\pi}{12} + \frac{\pi}{3} = \frac{5\pi}{12}$
   • For $n = 2$: $x = \frac{\pi}{12} + \frac{2\pi}{3} = \frac{9\pi}{12} = \frac{3\pi}{4}$
   • For $n = 3$: $x = \frac{\pi}{12} + \pi = \frac{13\pi}{12}$
   • For $n = 4$: $x = \frac{\pi}{12} + \frac{4\pi}{3} = \frac{17\pi}{12}$
   • For $n = 5$: $x = \frac{\pi}{12} + \frac{5\pi}{3} = \frac{21\pi}{12} = \frac{7\pi}{4}$

   All these values are in the interval $[0, 2\pi)$.

5. Answer: Thus, the solutions are: $x = \frac{\pi}{12}, \frac{5\pi}{12}, \frac{3\pi}{4}, \frac{13\pi}{12}, \frac{17\pi}{12}, \frac{7\pi}{4}$

So, the correct choice is A: $x = \frac{\pi}{12}, \frac{5\pi}{12}, \frac{3\pi}{4}, \frac{13\pi}{12}, \frac{17\pi}{12}, \frac{7\pi}{4}$.

Would you like further details or have any questions?

1. What are the general solutions to $\cot(3x) = 1$ in all intervals?
2. How does the period of cotangent affect the solutions in $[0, 2\pi)$?
3. Why is $x = \frac{\pi}{12} + \frac{n\pi}{3}$ the correct form for solutions?
4. How can we verify each solution is valid within $[0, 2\pi)$?
5. What are other methods to solve trigonometric equations?

Tip: Always check the interval of the solution range to ensure all values fit within it.