Question

If $\tan x+2\tan 2x+4\tan 4x+8\cot 8x=\sqrt{3}$ then the general solution of $x$ is

• A. $n\pi +\frac{\pi }{3},\forall n\in Z$
• B. $n\pi +\frac{\pi }{6},\forall n\in Z$
• C. $n\pi +\frac{\pi }{4},\forall n\in Z$
• D. $n\pi ,\forall n\in Z$

Answer 1

The correct option is B $n\pi +\frac{\pi }{6},\forall n\in Z$

Given,

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

or, $(\tan x-\cot x)+2\tan 2x+4\tan 4x+8\cot 8x+\cot x=\sqrt{3}$

or, $\tan x-\frac{1}{\tan x}+2\tan 2x+4\tan 4x+8\cot 8x+\cot x=\sqrt{3}$

or, $-2\times \frac{1-{\tan }^{2}x}{2\tan x}+2\tan 2x+4\tan 4x+8\cot 8x+\cot x=\sqrt{3}$

or, $-2\times \frac{1}{\tan 2x}+2\tan 2x+4\tan 4x+8\cot 8x+\cot x=\sqrt{3}$

or, $-2\cot 2x+2\tan 2x+4\tan 4x+8\cot 8x+\cot x=\sqrt{3}$

or, $-2\times \frac{1}{\tan 2x}+2\tan 2x+4\tan 4x+8\cot 8x+\cot x=\sqrt{3}$

or, $-4\times \frac{1-{\tan }^{2}2x}{2\tan 2x}+4\tan 4x+8\cot 8x+\cot x=\sqrt{3}$

or, $-4\cot 4x+4\tan 4x+8\cot 8x+\cot x=\sqrt{3}$

or, $-4\times \frac{1}{\tan 4x}+4\tan 4x+8\cot 8x+\cot x=\sqrt{3}$

or, $-8\times \frac{1-{\tan }^{2}4x}{2\tan 4x}+8\cot 8x+\cot x=\sqrt{3}$

or, $-8\cot 8x+8\cot 8x+\cot x=\sqrt{3}$

or, $\cot x=\sqrt{3}$

or, $\tan x=\frac{1}{\sqrt{3}}=\tan \frac{\pi }{6}$

or, $x=n\pi +\frac{\pi }{6}$. [ $n\in Z$]