Question

Solve the following equation:
${\cot }^{2}x-{\tan }^{2}x=8{\cot }^{2}2x$

Answer 1

${\cot }^{2}2x={\left(\cot 2x\right)}^2={\left(\frac{1-{\tan }^{2}x}{2\tan x}\right)}^2=\frac{1+{\tan }^{4}x-{2\tan }^{2}x}{{4\tan }^{2}x}$

${\cot }^{2}x-{\tan }^{2}x=\frac{2(1+{\tan }^{4}x-{2\tan }^{2}x)}{{\tan }^{2}x}$

$1+{\tan }^{4}x=2+{2\tan }^{4}x-{4\tan }^{2}x$

Let $\tan x=z.$

$1+z^4=2+{2z}^4-{4x}^2$

${3z}^4-{4z}^2+1=0$

$({3z}^2-1)(z^2-1)=0$

$\therefore$ $z=\pm 1$ and $z=\pm \frac{1}{\sqrt{3}}$

So, $x=n\pi \pm \frac{\pi }{4}$ and $x=x\pi \pm \frac{\pi }{6}$