Question

Solve the following equations.
$ta{n}^{2}xta{n}^{2}3xtan4x=ta{n}^{2}x-ta{n}^{2}3x+tan4x.$

Answer 1

$ta{n}^{2}xta{n}^{2}3x-tan4x=ta{n}^{2}x-ta{n}^{2}3x+tan4x$

$\therefore tan4x(ta{n}^{2}xta{n}^{2}3x-1)=ta{n}^{2}x-ta{n}^{2}3x$

$\therefore tan4x=\frac{ta{n}^{2}x-ta{n}^{2}3x}{ta{n}^{2}xta{n}^{2}3x-1}$

$=\frac{ta{n}^{2}3x-ta{n}^{2}x}{1-ta{n}^{2}xta{n}^{2}3x}$

$=\frac{(tan3x-tanx)(tan3x+tanx)}{(1-tanx-tan3x)(1+tanxtan3x)}$

$[\because a^2-b^2=(a+b\left)\right(a-b\left)\right]$

$\therefore tan4x=tan(3x-x)tan(3x+x)$

$\therefore tan4x=\left(tan2x\right)\left(tan4x\right)$

$\therefore tan4x(tan2x-1)=0$

$\therefore tan4x=0$

$\therefore 4x=n\pi$

$\therefore x=\frac{n\pi }{4}$

$tan2x-1=0$

$\therefore tan2x=1$

$tan2x=tan\pi /4$

$\therefore 2x=n\pi +\pi /4$ (general solution)

$\therefore x=\frac{n\pi }{2}+\frac{\pi }{8}$