Answer :


Ideas required to solve the problem:


The general solution of any trigonometric equation is given as –


• sin x = sin y, implies x = nπ + (– 1)ny, where n Z.


• cos x = cos y, implies x = 2nπ ± y, where n Z.


• tan x = tan y, implies x = nπ + y, where n Z.


given,


tan x + tan 2x + tan 3x = 0


In order to solve the equation we need to reduce the equation into factor form so that we can equate the ratios with 0 and can solve the equation easily


As if we expand tan 3x = tan ( x + 2x) we will get tan x + tan 2x common.


tan x + tan 2x + tan 3x = 0


tan x + tan 2x + tan (x + 2x) = 0


As, tan (A + B) =


tan x + tan 2x +




tan x + tan 2x = 0 or 2 – tan x tan 2x = 0


Using, tan 2x = we have,


tan x = tan (-2x) or 2 –


tan x = tan(-2x) or 2 – 4tan2 x = 0 tan x = 1/ √2


Let 1/√2 = tan α and if tan x = tan y, implies x = nπ + y, where n Z


x = nπ + (-2x) or tan x = tan α x = mπ + α


3x = nπ or x = mπ + α



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