Question

In (0, π), the number of solutions of the equation ​ \[\tan x + \tan 2x + \tan 3x = \tan x \tan 2x \tan 3x\] is 

Options

• 7

• 5

• 4

• 2

Answer 1

2
Given equation:
\[\tan x + \tan2x + \tan3x = \tan x \tan2x \tan3x\]
\[ \Rightarrow \tan x + \tan2x = - \tan3x + \tan x \tan2x \tan3x\]
\[ \Rightarrow \tan x + \tan2x = - \tan3x (1 - \tan x \tan2x)\]
\[ \Rightarrow \frac{\tan x + \tan2x}{1 - \tan x \tan 2x} = - \tan3x\]
\[ \Rightarrow \tan ( x + 2x) = - \tan3x\]
\[ \Rightarrow \tan3x = - \tan3x\]
\[ \Rightarrow 2 \tan3x = 0\]
\[ \Rightarrow \tan3x = 0\]
\[ \Rightarrow 3x = n\pi\]
\[ \Rightarrow x = \frac{n\pi}{3}\]
Now,
\[x = \frac{\pi}{3} , n = 1\]
\[x = \frac{2\pi}{3} , n = 2\]
\[x = \frac{3\pi}{3} = 180^\circ\], which is not possible, as it is not in the interval \[(0, 2\pi)\].
Hence, the number of solutions of the given equation is 2.