Sum

Solve the following equation:

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

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#### Solution

Given:

\[\tan3x + \tan x = 2 \tan2x\]

Now,

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

\[ \Rightarrow \tan x (1 + \tan3x \tan2x) = \tan x(1 + \tan2x \tan x) \left[ \tan \left( A - B \right) = \frac{\tan A - \tan B}{1 + \tan A \tan B} \right] \]

\[ \Rightarrow \tan x (1 + \tan3x\tan2x - 1 - \tan2x \tan x) = 0\]

\[ \Rightarrow \tan x \tan2x (\tan3x - \tan x) = 0\]

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

\[\tan x = 0\] or,

\[\tan3x - \tan x = 0\]

And,

\[\tan 2x = 0 \Rightarrow 2x = n\pi \Rightarrow x = \frac{n\pi}{2}, n \in Z\]

or,

\[\tan 3x - \tan x = 0 \Rightarrow \tan 3x = \tan x \Rightarrow 3x = n\pi + x \Rightarrow 2x = n\pi \Rightarrow x = \frac{n\pi}{2}, n \in Z\]

And,

\[\tan x = 0 \Rightarrow x = m\pi, m \in Z\]

∴ \[x = \frac{n\pi}{2}, n \in Z\] or

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

Concept: Trigonometric Equations

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