Question

Solve the following equation:

\[\tan x + \tan 2x + \tan 3x = 0\]

Answer 1

\[\tan x + \tan 2x + \tan 3x = 0\]
Now,

\[\tan x + \tan2x + \tan (x + 2x) = 0\]
\[ \Rightarrow \tan x + \tan2x + \left( \frac{\tan x + \tan 2x}{1 - \tan x \tan 2x} \right) = 0\]
\[ \Rightarrow (\tan x + \tan2x) (1 - \tan x\tan2x) + \tan x + \tan2x = 0\]
\[ \Rightarrow (\tan x + \tan2x) (2 - \tan x \tan2x) = 0\]

\[\Rightarrow \tan x + \tan 2x = 0\] or

\[2 - \tan x \tan2x = 0\]

Now,

\[\tan x + \tan2x = 0 \]

\[ \Rightarrow \tan x = - \tan2x\]

\[ \Rightarrow \tan x = \tan - 2x\]

\[ \Rightarrow x = n\pi - 2x \]

\[ \Rightarrow 3x = n\pi \]

\[ \Rightarrow x = \frac{n\pi}{3}, n \in Z\]

And,

\[2 - \tan x \tan2x = 0 \]
\[ \Rightarrow \tan x \tan2x = 2 \]
\[ \Rightarrow \frac{\sin x}{\cos x}\frac{\sin2x}{\cos2x} = 2\]
\[ \Rightarrow \frac{2 \sin^2 x \cos x}{\cos x} = 2 \cos^2 x - 2 \sin^2 x\]
\[ \Rightarrow 4 \sin^2 x = 2 \cos^2 x \]
\[ \Rightarrow \tan^2 x = \frac{1}{2} \Rightarrow \tan^2 x = \tan^2 \alpha \]
\[ \Rightarrow x = m\pi + \alpha, m \in Z, \alpha = \tan^{- 1} \left( \frac{1}{2} \right)\]

∴ \[x = \frac{n\pi}{3}, n \in Z\] or

\[x = m\pi + \alpha, m \in Z\]

Here,

\[x = m\pi + \alpha, m \in Z\]