The general solution of the trigonometric equation $tan x + tan 2 x + tan 3 x = tan x . tan 2 x . tan 3 x$ is

x = n π

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n π \pm \frac{π}{3}

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x = 2 n π

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x = \frac{n π}{3}

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Solution

The correct option is B $x = \frac{n π}{3}$
$tan x + tan 2 x + tan 3 x = tan x tan 2 x tan 3 x$

$tan x + tan 2 x = - tan 3 x + tan x tan 2 x tan 3 x$

$tan x + tan 2 x = - tan 3 x (1 - tan x tan 2 x)$

$\frac{tan x + tan 2 x}{1 - tan x tan 2 x} = - tan 3 x$

$tan (x + 2 x) = - tan 3 x$

$tan 3 x = - tan 3 x$

$2 tan 3 x = 0$

$tan 3 x = 0$

$3 x = n π, n \in Z$

$x = \frac{n π}{3}, n \in Z$

Suggest Corrections

Similar questions

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

(where n \in Z)

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

f (x) = tan x - {tan}^{2} x + {tan}^{3} x, x \neq n π + \frac{π}{2}

t a n^{2} θ = 3 i s (n ϵ Z)

x

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

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