Advertisement Remove all ads

Solve the Following Equation: Tan X + Tan 2 X = Tan 3 X - Mathematics

Sum

Solve the following equation:

\[\tan x + \tan 2x = \tan 3x\]
Advertisement Remove all ads

Solution

Given:

\[\tan x + \tan2x = \tan3x\]
Now, 
\[\tan x + \tan2x = \tan (x + 2x)\]
\[ \Rightarrow \tan x + \tan 2x = \frac{\tan x + \tan2x}{1 - \tan x \tan2x}\]
\[ \Rightarrow \tan x + \tan2x - \frac{\tan x + \tan2x}{1 - \tan x \tan2x} = 0\]
\[ \Rightarrow (\tan x + \tan2x) (1 - \tan x \tan2x) - (\tan x + \tan2x) = 0\]
\[ \Rightarrow (\tan x + \tan 2x) (1 - \tan x \tan2x - 1) = 0\]
\[ \Rightarrow (\tan x + \tan2x) ( - \tan x \tan2x) = 0\]
\[\Rightarrow \tan x + \tan 2x = 0\] 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

\[x = \frac{n\pi}{3}, n \in Z\]
  Is there an error in this question or solution?
Advertisement Remove all ads

APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 11 Trigonometric equations
Exercise 11.1 | Q 5.2 | Page 22
Advertisement Remove all ads

Video TutorialsVIEW ALL [1]

Advertisement Remove all ads
Share
Notifications

View all notifications


      Forgot password?
View in app×