Question

Solve the following equation:
$\tan x+\tan 2x+\tan 3x=0$

Answer 1

$tanx+tan2x+tan3x=\mathrm{0...}\left(1\right)$

we have $tan3x=tan(x+2x)=(tanx+tan2x)(1-tanx.tan2x)$

$\Rightarrow tanx+tan2x=tan3x-tanx.tan2x.tanx3x$

using the above realation equation (1) can written as

$\Rightarrow tan3x-tanx.tan2x.tan3x+tan3x=0$

$\Rightarrow tan3x(2-tanx.tan2x)=0$

$\therefore tan3x=0$

$x=x\pi /3;xϵ$ integers.

and, $2-tanx.tan2x=0$

$\Rightarrow tanx.tan2x=2$

$\Rightarrow tanx\frac{2tanx}{1-ta{n}^{2}x}=x$

$\Rightarrow ta{n}^{2}x=1-ta{n}^{2}x$

$\Rightarrow 2ta{n}^{2}x=1$

$\Rightarrow tanx=\pm 1/\sqrt{2}$

$x=n\pi \pm ta{n}^{-1}(\pm 1/\sqrt{2})$