Question

The number of solution of the equation
$tanx+tan2x+tan3x=tanx.tan2x.tan3x$ in the interval $\left[0,\pi \right]$

Answer 1

We know,

$\tan 3x=\frac{\tan x+\tan 2x}{1-\tan x\tan 2x}$

$\Rightarrow \tan 3x-\tan 3x\tan x\tan 2x=\tan x+\tan 2x$

$\Rightarrow \tan 3x-\tan x-\tan 2x=\tan x\tan 2x\tan 3x$

From the given equation,

$\Rightarrow \tan x+\tan 2x+\tan 3x=\tan x\tan 2x\tan 3x$

$\therefore \tan 3x-\tan x-\tan 2x=\tan x+\tan 2x+\tan 3x$

$\Rightarrow -2\tan x=2\tan 2x$

$\Rightarrow -\tan x=\frac{2\tan x}{1-\tan {}^{2}x}$

$\Rightarrow 1-\tan {}^{2}x=-2$

$\Rightarrow \tan {}^{2}x=3$

$\Rightarrow \tan x=\pm \sqrt{3}$

$\therefore Solutionsare\frac{\pi }{3},\frac{2\pi }{3}$

Hence, the number of solutions is 2.