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Question: If tan x + tan 2x + tan 3x = 0, then x =? \( {\text{A}}{\text{. n}}\pi {\text{ + }}\dfrac{\pi ...

If tan x + tan 2x + tan 3x = 0, then x =?
A. nπ + π3 B. nπ+π4 C. nπ or nπ±π3 D. 2nπ  {\text{A}}{\text{. n}}\pi {\text{ + }}\dfrac{\pi }{3} \\\ {\text{B}}{\text{. n}}\pi + \dfrac{\pi }{4} \\\ {\text{C}}{\text{. n}}\pi {\text{ or n}}\pi \pm \dfrac{\pi }{3} \\\ {\text{D}}{\text{. 2n}}\pi \\\

Explanation

Solution

To find the value of x, we rearrange the given equation and convert it in terms of sin and cos functions. Then we apply the formulae of sin (a + b) and cos (a + b) and simplify.

Complete step-by-step answer :
Given data, tan x + tan 2x + tan 3x = 0
⟹tan x + tan 2x = -tan 3x
sin xcos x+sin 2xcos 2x=sin 3xcos 3x\Rightarrow \dfrac{{{\text{sin x}}}}{{{\text{cos x}}}} + \dfrac{{{\text{sin 2x}}}}{{{\text{cos 2x}}}} = - \dfrac{{{\text{sin 3x}}}}{{{\text{cos 3x}}}}
sin x cos 2x + cos x sin 2xcos x cos 2x=sin 3xcos 3x\Rightarrow \dfrac{{{\text{sin x cos 2x + cos x sin 2x}}}}{{{\text{cos x cos 2x}}}} = - \dfrac{{{\text{sin 3x}}}}{{{\text{cos 3x}}}}
We know the function Sin (a + b) is given as (Sin a Cos b + Cos a Sin b).
Comparing with the above equation, here a = x and b = 2x.
sin (2x + x) cos 3x = - cos x cos 2x sin 3x\Rightarrow {\text{sin }}\left( {{\text{2x + x}}} \right){\text{ cos 3x = - cos x cos 2x sin 3x}}
sin 3x cos 3x + cos x cos 2x sin 3x = 0\Rightarrow {\text{sin 3x cos 3x + cos x cos 2x sin 3x = 0}}
sin 3x (cos 3x + cos x cos 2x) = 0\Rightarrow {\text{sin 3x }}\left( {{\text{cos 3x + cos x cos 2x}}} \right){\text{ = 0}}
sin 3x (cos (2x + x) + cos x cos 2x) = 0\Rightarrow {\text{sin 3x }}\left( {{\text{cos }}\left( {{\text{2x + x}}} \right){\text{ + cos x cos 2x}}} \right){\text{ = 0}}
We know the function Cos (a + b) is given as (Cos a Cos b - Sin a Sin b).
Comparing with the above equation, here a = x and b = 2x
sin 3x (cos x cos 2x - sin x sin 2x + cos x cos 2x)= 0\Rightarrow {\text{sin 3x }}\left( {{\text{cos x cos 2x - sin x sin 2x + cos x cos 2x}}} \right) = {\text{ 0}}
 - sin 3x sin x sin 2x = 0\Rightarrow {\text{ - sin 3x sin x sin 2x = 0}}
sin x sin 2x sin 3x = 0\Rightarrow {\text{sin x sin 2x sin 3x = 0}}

This means either of the terms is zero.
Either, sin 3x = 0
i.e., 3x = nπ, n ε I
i.e., x =nπ3\dfrac{{{\text{n}}\pi }}{3}, n ε I

Or, sin 2x = 0
i.e., 2x = nπ, n ε I
i.e., x =nπ2\dfrac{{{\text{n}}\pi }}{2}, n ε I
Or, sin x = 0
i.e., nπ, n ε I
But putting x =nπ2\dfrac{{{\text{n}}\pi }}{2}, does not satisfy the equation. (E.g. Putting x =nπ2\dfrac{{{\text{n}}\pi }}{2}, tanπ2\dfrac{\pi }{2}= undefined)
Hence the solution set is x = nπ and x =nπ3\dfrac{{{\text{n}}\pi }}{3}
Hence Option C is the correct answer.

Note : In order to solve this type of problems the key is to have adequate knowledge in trigonometric formulae such as Sin (a + b) and Cos (a – b) and tan θ =Sin θCos θ\dfrac{{{\text{Sin }}\theta }}{{{\text{Cos }}\theta }}. It is important to identify that ifsin x sin 2x sin 3x = 0{\text{sin x sin 2x sin 3x = 0}}, then either of them is zero. Also Sin π = 0. And the letter I represents the set of integers here.