The general solution of the equation $tan x + tan 2 x + tan x tan 2 x = 1$ is

x = \frac{n π}{3} + \frac{π}{2}

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x = \frac{n π}{3} + \frac{π}{12}

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x = \frac{n π}{3} + \frac{π}{1}

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x = \frac{n π}{3} + \frac{1}{12}

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Solution

The correct option is D $x = \frac{n π}{3} + \frac{π}{12}$
$t a n x + t a n 2 x + t a n x t a n 2 x = 1 \Rightarrow t a n x + t a n 2 x = 1 - t a n x t a n 2 x \Rightarrow \frac{t a n x + t a n 2 x}{1 - t a n x t a n 2 x} = 1 \Rightarrow t a n 3 x = 1 3 x = n π + \frac{π}{4} x = \frac{n π}{3} + \frac{π}{12}$

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Q. The general solution of

tan x + tan 2 x + \sqrt{3} tan x \cdot tan 2 x = \sqrt{3}

(where n \in Z)

The general solution of $t a n 2 x = - c o t (x + \frac{π}{3}) i s (n ϵ Z)$

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

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The general solution of the equation tanx+tan2x+tanxtan2x=1 is

The correct option is D x=nπ3+π12tanx+tan2x+tanxtan2x=1⇒tanx+tan2x=1−tanxtan2x⇒tanx+tan2x1−tanxtan2x=1⇒tan3x=13x=nπ+π4x=nπ3+π12

The general solution of the equation $tan x + tan 2 x + tan x tan 2 x = 1$ is