The general solution of tan x-tan 2 x-3tan x -tan 2 x-3 iswhere n - Z A- 2 n -1-6B- n -9-3C- n -3-9D- n -3-1 n -6

The correct option is C nπ3+π 9Given: 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 n ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

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}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

2 n π \pm \frac{π}{6}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{n π}{9} + \frac{π}{3}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{n π}{3} + \frac{π}{9}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

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 $3 x = n π + \frac{π}{3} \Rightarrow x = \frac{n π}{3} + \frac{π}{9}, n \in Z$

Suggest Corrections

Similar questions

Q. The general solution of

tan α + 2 tan 2 α + 4 tan 4 α + 8 cot 8 α = \sqrt{3}

(where n \in Z)

Q. The general solution of

t a n^{2} θ = 3 i s (n ϵ Z)

Q. The general solution of

(\sqrt{3} - 1) sin θ + (\sqrt{3} + 1) cos θ = 2

is
(where

n \in Z

)

$The general value of x satisfying the equation \sqrt{3} s i n x + c o s x = \sqrt{3} i s g i v e n b y$

Q. General solution of

x

for which

{cos}^{2} x = \frac{sin x tan x}{6}

is :

Join BYJU'S Learning Program

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
$(where n \in Z)$