solve the following equation on the interval 0,2π).\n\ncot(3x)=\\frac{\\sqrt{3}}{3}\n\nselect the correct…

solve the following equation on the interval 0,2π).\n\ncot(3x)=\\frac{\\sqrt{3}}{3}\n\nselect the correct choice below and, if necessary, fill in the answer box to complete your choice.\n\na. x=□\n\n(type an exact answer, using π as needed. use a comma to separate answers as needed. type your answer in radians. simplify your answer. use integers or fractions for any numbers in the expression.)\n\nb. there is no solution.
Answer
Explanation:
Step1: Recall the value of cotangent
We know that (\cot\theta=\frac{\sqrt{3}}{3}) when (\theta = \frac{\pi}{3}+n\pi), (n\in\mathbb{Z}). So for (\cot(3x)=\frac{\sqrt{3}}{3}), we have (3x=\frac{\pi}{3}+n\pi).
Step2: Solve for (x)
Divide both sides of (3x=\frac{\pi}{3}+n\pi) by (3). Then (x = \frac{\pi}{9}+\frac{n\pi}{3}).
Step3: Find solutions in the interval ([0,2\pi))
When (n = 0), (x=\frac{\pi}{9}). When (n = 1), (x=\frac{\pi}{9}+\frac{\pi}{3}=\frac{4\pi}{9}). When (n = 2), (x=\frac{\pi}{9}+\frac{2\pi}{3}=\frac{7\pi}{9}).
When (n = 3), (x=\frac{\pi}{9}+\pi=\frac{10\pi}{9}). When (n = 4 ), (x=\frac{\pi}{9}+\frac{4\pi}{3}=\frac{13\pi}{9}). When (n = 5), (x=\frac{\pi}{9}+\frac{5\pi}{3}=\frac{16\pi}{9}).
Answer:
(x=\frac{\pi}{9},\frac{4\pi}{9},\frac{7\pi}{9},\frac{10\pi}{9},\frac{13\pi}{9},\frac{16\pi}{9})