Question

The general solution of the equation \( \tan x + \tan 2x - \tan 3x = 0 \) is:

• A. \( \{ x | x = n\pi \pm \frac{\pi}{3} \text{ or } \frac{n\pi}{2}, n \in \mathbb{Z} \} \)
• B. \( \{ x | x = n\pi \pm \frac{\pi}{3} \text{ or } n\pi, n \in \mathbb{Z} \} \)
• C. \( \{ x | x = n\pi \pm \frac{\pi}{3} \text{ or } \frac{n\pi}{2}, n \in \mathbb{Z} \} \)
• D. \( \{ x | x = n\pi \pm \frac{\pi}{6} \text{ or } \frac{n\pi}{2}, n \in \mathbb{Z} \} \)

Hint:

For solving equations involving tangent, always check for standard identities like the sum and triple angle formulas.

Answer 1

The Correct Option is B

We are required to solve the given equation: \[ \tan x + \tan 2x - \tan 3x = 0 \] Step 1: Use the Identity for \( \tan 3x \) From the identity: \[ \tan 3x = \frac{\tan x + \tan 2x}{1 - \tan x \tan 2x} \] Substituting this into the original equation: \[ \tan x + \tan 2x - \frac{\tan x + \tan 2x}{1 - \tan x \tan 2x} = 0 \] Step 2: Combine Like Terms Multiply through by \( 1 - \tan x \tan 2x \) to eliminate the denominator: \[ (\tan x + \tan 2x)(1 - \tan x \tan 2x) - (\tan x + \tan 2x) = 0 \] Simplifying, \[ (\tan x + \tan 2x) (1 - \tan x \tan 2x) - (\tan x + \tan 2x) = 0 \] Factoring out \( (\tan x + \tan 2x) \), \[ (\tan x + \tan 2x) \left[ (1 - \tan x \tan 2x) - 1 \right] = 0 \] \[ (\tan x + \tan 2x)(-\tan x \tan 2x) = 0 \] Step 3: Identify the Conditions - \( \tan x + \tan 2x = 0 \) - \( \tan x \tan 2x = 0 \) Step 4: Solving Each Condition For \( \tan x + \tan 2x = 0 \) \[ \tan x = -\tan 2x \] Using the identity: \[ \tan x = -\tan \left(\pi - 2x \right) \] Thus, \[ x = n\pi \pm \frac{\pi}{3} \] For \( \tan x \tan 2x = 0 \) This implies: \[ \tan x = 0 \quad \text{or} \quad \tan 2x = 0 \] - \( \tan x = 0 \) ⇒ \( x = n\pi \) - \( \tan 2x = 0 \) ⇒ \( x = \frac{n\pi}{2} \) Step 5: Combine Solutions The general solution is: \[ x = n\pi \pm \frac{\pi}{3} \quad \text{or} \quad x = n\pi \] Step 6: Final Answer

\[Correct Answer: (2) \ \{ x \mid x = n\pi \pm \frac{\pi}{3} \text{ or } n\pi, n \in \mathbb{Z} \}\]

Answer 2

To find the general solution of the equation \( \tan x + \tan 2x - \tan 3x = 0 \), let's analyze it step by step:

We start by using the identity for the tangent of a sum to simplify \(\tan 3x\):

\[\tan 3x = \frac{\tan x + \tan 2x}{1 - \tan x \tan 2x}\]

This equation implies:

\[ \frac{\tan x + \tan 2x}{1 - \tan x \tan 2x} = \tan 3x\]

Substituting in our original equation gives:

\[\tan x + \tan 2x - \frac{\tan x + \tan 2x}{1 - \tan x \tan 2x} = 0\]

Combine the terms over a common denominator:

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

Combine and simplify:

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

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

So, either \(\tan x + \tan 2x = 0\) or \(\tan x \tan 2x = 0\):

1) For \(\tan x + \tan 2x = 0\), \(\tan x = -\tan 2x\).

Using the identity, \(\tan 2x = \frac{2\tan x}{1-\tan^2 x}\):

\[\tan x = -\frac{2\tan x}{1-\tan^2 x}\]

which leads to: \(-\tan^2 x = 2\). As \(\tan^2 x = -2\) is impossible in real numbers, it gives no real root.

2) \(\tan x \tan 2x = 0\) implies \(\tan x = 0\) or \(\tan 2x = 0\).

For \(\tan x = 0\), \(x = n\pi, n \in \mathbb{Z}\).

For \(\tan 2x = 0\), \(2x = m\pi \Rightarrow x = \frac{m\pi}{2}, m \in \mathbb{Z}.\)

However, the condition for \(\tan x = -\tan 2x\) does give us the specific solutions \(\tan x = \tan(-2x)\), which further imply: \(x = n\pi \pm \frac{\pi}{3}\).

Thus, the general solution is:

\(\{ x | x = n\pi \pm \frac{\pi}{3} \text{ or } n\pi, n \in \mathbb{Z} \}\).