Sum

Solve the following equation:

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

Advertisement Remove all ads

#### Solution

Given:

\[\tan x + \tan2x = \tan3x\]

Now,

\[\tan x + \tan2x = \tan (x + 2x)\]

\[ \Rightarrow \tan x + \tan 2x = \frac{\tan x + \tan2x}{1 - \tan x \tan2x}\]

\[ \Rightarrow \tan x + \tan2x - \frac{\tan x + \tan2x}{1 - \tan x \tan2x} = 0\]

\[ \Rightarrow (\tan x + \tan2x) (1 - \tan x \tan2x) - (\tan x + \tan2x) = 0\]

\[ \Rightarrow (\tan x + \tan 2x) (1 - \tan x \tan2x - 1) = 0\]

\[ \Rightarrow (\tan x + \tan2x) ( - \tan x \tan2x) = 0\]

\[ \Rightarrow \tan x + \tan 2x = \frac{\tan x + \tan2x}{1 - \tan x \tan2x}\]

\[ \Rightarrow \tan x + \tan2x - \frac{\tan x + \tan2x}{1 - \tan x \tan2x} = 0\]

\[ \Rightarrow (\tan x + \tan2x) (1 - \tan x \tan2x) - (\tan x + \tan2x) = 0\]

\[ \Rightarrow (\tan x + \tan 2x) (1 - \tan x \tan2x - 1) = 0\]

\[ \Rightarrow (\tan x + \tan2x) ( - \tan x \tan2x) = 0\]

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

\[\tan x \tan2x = 0\]

Now,

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

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

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

\[ \Rightarrow x = n\pi - 2x, n \in Z\]

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

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

And,

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

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

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

\[ \Rightarrow x = n\pi - 2x, n \in Z\]

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

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

∴ \[x = \frac{n\pi}{3}, n \in Z\] or

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

Concept: Trigonometric Equations

Is there an error in this question or solution?

Advertisement Remove all ads

#### APPEARS IN

Advertisement Remove all ads

Advertisement Remove all ads