The general solution of-tan x-tan 2x-3-tan x -tan 2 x - -3- is nbsp-where n -

Given: tan x+tan 2x+√(3)tan x ·tan 2 x = √(3) ⇒tan x+tan 2x nbsp;= √(3) (1 tan x ·tan 2 x) ⇒tan x+tan 2x1 tan x ·tan 2 x nbsp; = √(3) ⇒tan 3x = tanπ3 So,g ...

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Standard XII

Mathematics

General Solution of Trigonometric Equation

The general s...

Question

The general solution of $tan x + tan 2 x + \sqrt{3} tan x \cdot tan 2 x = \sqrt{3}$ is
$(where n \in Z)$

\frac{n π}{3} + (- 1)^{n} \frac{π}{6}

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2 n π \pm \frac{π}{6}

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\frac{n π}{9} + \frac{π}{3}

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\frac{n π}{3} + \frac{π}{9}

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Solution

The correct option is D $\frac{n π}{3} + \frac{π}{9}$
Given:
$tan x + tan 2 x + \sqrt{3} tan x \cdot tan 2 x = \sqrt{3} \Rightarrow tan x + tan 2 x = \sqrt{3} (1 - tan x \cdot tan 2 x) \Rightarrow \frac{tan x + tan 2 x}{1 - tan x \cdot tan 2 x} = \sqrt{3} \Rightarrow tan 3 x = tan \frac{π}{3}$
So,general solution is $3 x = n π + \frac{π}{3} \Rightarrow x = \frac{n π}{3} + \frac{π}{9}, n \in Z$

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The general solution of tanx+tan2x+√3tanx⋅tan2x=√3 is (where n∈Z)

The correct option is D nπ3+π9Given: tanx+tan2x+√3tanx⋅tan2x=√3⇒tanx+tan2x=√3(1−tanx⋅tan2x)⇒tanx+tan2x1−tanx⋅tan2x=√3⇒tan3x=tanπ3 So,general solution is 3x=nπ+π3⇒x=nπ3+π9, n∈Z

The general solution of $tan x + tan 2 x + \sqrt{3} tan x \cdot tan 2 x = \sqrt{3}$ is
$(where n \in Z)$