Question

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

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

Answer 1

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

$\tan x+2\tan 2x+4\tan 4x+8\cot 8x=\sqrt{3}$
$formulaused$
$\tan x-\cot x$
$=\tan x-\frac{1}{\tan x}$
$=\frac{{\tan }^{2}x-1}{\tan x}$
$=-2\left[\frac{1-{\tan }^{2}x}{2\tan x}\right]$
$=-2\left[\frac{1}{\frac{2\tan x}{1-{\tan }^{2}x}}\right]$
$=-2\left[\frac{1}{\tan 2x}\right]$
$=-2\cot x$
$\tan x=\tan \alpha$
$x=n\pi +\alpha$
$\tan x+2\tan 2x+4\tan 4x+8\cot 8x=\sqrt{3}$
$addingandsubtracting\cot x$
$\tan x-\cot x+2\tan 2x+4\tan 4x+8\cot 8x+\cot x=\sqrt{3}$
$\Rightarrow -2\cot 2x+2\tan 2x+4\tan 4x+8\cot 8x+\cot x=\sqrt{3}$
$\Rightarrow 2[\tan 2x-\cot 2x]+4\tan 4x+8\cot 8x+\cot x=\sqrt{3}$
$\Rightarrow 2[-2\cot 4x]+4\tan 4x+8\cot 8x+\cot x=\sqrt{3}$
$\Rightarrow -4\cot 4x+4\tan 4x+8\cot 8x+\cot x=\sqrt{3}$
$\Rightarrow 4[\tan 4x-\cot 4x]+8\cot 8x+\cot x=\sqrt{3}$
$\Rightarrow 4[-2\cot 8x]+8\cot 8x+\cot x=\sqrt{3}$
$\Rightarrow -8\cot x+8\cot x+\cot x=\sqrt{3}$
$\Rightarrow \cot x=\sqrt{3}$
$\Rightarrow \frac{1}{\tan x}=\sqrt{3}$
$\Rightarrow \tan x=\frac{1}{\sqrt{3}}=\tan \left(\frac{\pi }{6}\right)$
$x=n\pi +\frac{\pi }{6}$