Answer :

tan x+tan2x+tan3x -tan xtan2xtan3x=0


tan x + tan2x + tan3x (1-tan xtan2x) = 0


tan x + tan2x = -tan3x (1-tan xtan2x)



tan 3x=-tan3x


2 tan3x=0


tan 3x=0


3x=2nπ



For


n=0, x=0


n=1,





so, there are only two possible solutions


Option D

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