$\tan 3x=\frac{\sqrt{3} }{3} =\frac{1}{\sqrt{3} } =tan\frac{\pi }{6} =tan (n\pi +\frac{\pi }{6} )\\3x=n\pi +\frac{\pi }{6} =\frac{(6n+1)\pi }{6} \\x=\frac{(6n+1)\pi }{18} \\where~x~is~an~integer.$

Step-by-step explanation:

tan 3x=\frac{\sqrt{3} }{3} =\frac{1}{\sqrt{3} } =tan\frac{\pi }{6} =tan (n\pi +\frac{\pi }{6} )\\3x=n\pi +\frac{\pi }{6} =\frac{(6n+1)\pi }{6} \\x=\frac{(6n+1)\pi }{18} \\

where~x~is~an~integer.

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