Question

Which of the following is the total general solution of the equation $4\cot 2x={\cot }^{2}x-{\tan }^{2}x$ ?

Answer 1

The given equation is $4\cot 2x={\cot }^{2}x-{\tan }^{2}x$
$\Rightarrow \frac{4}{\tan 2x}=\frac{1}{{\tan }^{2}x}-{\tan }^{2}x$

Using $\tan 2\theta =\frac{2\tan \theta }{1-{\tan }^2\theta }$

$\Rightarrow \frac{4(1-{\tan }^{2}x)}{2\tan x}=\frac{1-{\tan }^{4}x}{{\tan }^{2}x}$
$\Rightarrow 2{\tan }^{2}x(1-{\tan }^{2}x)=\tan x(1-{\tan }^{4}x)$
Using $(a^4-b^4)=(a^2-b^2)(a^2+b^2)$
$\Rightarrow 2{\tan }^{2}x(1-{\tan }^{2}x)=\tan x(1-{\tan }^{2}x)(1+{\tan }^{2}x)$
$\Rightarrow 2{\tan }^{2}x(1-{\tan }^{2}x)-\tan x(1-{\tan }^{2}x)(1+{\tan }^{2}x)=0$
$\Rightarrow \tan x(1-{\tan }^{2}x)[2\tan x-(1+{\tan }^{2}x\left)\right]=0$
$\Rightarrow \tan x(1-{\tan }^{2}x)\left[\right({\tan }^{2}x-2\tan x+1\left)\right]=0$
$\Rightarrow \tan x(1-{\tan }^{2}x)(\tan x-1{)}^2=0$

$\Rightarrow \tan x=0$ or $\tan x=\pm 1$
$\Rightarrow \tan x=\tan 0$ or $\tan x=\pm \tan \frac{\pi }{4}$
Using
$\tan \theta =\tan \alpha \Rightarrow \theta =n\pi +\alpha ,n\in Z$

$\Rightarrow x\in \{n\pi ,n\in Z\}\cup \{n\pi \pm \frac{\pi }{4},n\in Z\}$