Question
tan3x tan2x tanx = tan3x - tan2x-tanx
Answers
GPT-4o mini
To solve the equation \( \tan 3x \tan 2x \tan x = \tan 3x - \tan 2x - \tan x \), we can start by reorganizing the equation:
\[
\tan 3x \tan 2x \tan x + \tan 2x + \tan x - \tan 3x = 0
\]
This equation can be quite complicated, so let's try to evaluate it by using some identities, particularly the tangent angle addition formulas.
Recall that:
1. \( \tan 2x = \frac{2 \tan x}{1 - \tan^2 x} \)
2. \( \tan 3x = \frac{3 \tan x - \tan^3 x}{1 - 3 \tan^2 x} \)
Substituting these identities into the original equation would be complex, so let's look for solutions by checking specific values of \( x \).
### Testing Specific Values
1. **Check \( x = 0 \)**:
\[
\tan 0 = 0 \Rightarrow LHS = 0 \cdot 0 \cdot 0 = 0
\]
\[
RHS = 0 - 0 - 0 = 0
\]
So, \( x = 0 \) is a solution.
2. **Check \( x = \frac{\pi}{4} \)**:
\[
\tan \frac{\pi}{4} = 1 \Rightarrow \tan 2x = \tan \frac{\pi}{2} = \text{undefined} \quad \text{ (not a valid check)}
\]
3. **Check \( x = \frac{\pi}{6} \)**:
\[
\tan \frac{\pi}{6} = \frac{1}{\sqrt{3}}, \quad \tan 2x = \tan \frac{\pi}{3} = \sqrt{3}, \quad \tan 3x = \tan \frac{\pi}{2} = \text{undefined} \quad \text{ (not a valid check)}
\]
4. **Check \( x = \frac{\pi}{3} \)**:
\[
\tan \frac{\pi}{3} = \sqrt{3}, \quad \tan 2x = \tan \frac{2\pi}{3} = -\sqrt{3}, \quad \tan 3x = \tan \pi = 0
\]
\[
LHS = 0 \cdot (-\sqrt{3}) \cdot \sqrt{3} = 0
\]
\[
RHS = 0 - (-\sqrt{3}) - \sqrt{3} = 0
\]
So, \( x = \frac{\pi}{3} \) is also a solution.
### Conclusion
The solutions we find are \( x = 0 \) and \( x = \frac{\pi}{3} \). To find more solutions, it may involve analyzing the periodicity of the tangent function and solving for increments of \( \pi \).
Thus, the complete solution will be of the form:
\[
x = n\pi \quad \text{for } n \in \mathbb{Z} \quad \text{and} \quad x = \frac{\pi}{3} + k\pi \quad \text{for } k \in \mathbb{Z}.
\]
\[
\tan 3x \tan 2x \tan x + \tan 2x + \tan x - \tan 3x = 0
\]
This equation can be quite complicated, so let's try to evaluate it by using some identities, particularly the tangent angle addition formulas.
Recall that:
1. \( \tan 2x = \frac{2 \tan x}{1 - \tan^2 x} \)
2. \( \tan 3x = \frac{3 \tan x - \tan^3 x}{1 - 3 \tan^2 x} \)
Substituting these identities into the original equation would be complex, so let's look for solutions by checking specific values of \( x \).
### Testing Specific Values
1. **Check \( x = 0 \)**:
\[
\tan 0 = 0 \Rightarrow LHS = 0 \cdot 0 \cdot 0 = 0
\]
\[
RHS = 0 - 0 - 0 = 0
\]
So, \( x = 0 \) is a solution.
2. **Check \( x = \frac{\pi}{4} \)**:
\[
\tan \frac{\pi}{4} = 1 \Rightarrow \tan 2x = \tan \frac{\pi}{2} = \text{undefined} \quad \text{ (not a valid check)}
\]
3. **Check \( x = \frac{\pi}{6} \)**:
\[
\tan \frac{\pi}{6} = \frac{1}{\sqrt{3}}, \quad \tan 2x = \tan \frac{\pi}{3} = \sqrt{3}, \quad \tan 3x = \tan \frac{\pi}{2} = \text{undefined} \quad \text{ (not a valid check)}
\]
4. **Check \( x = \frac{\pi}{3} \)**:
\[
\tan \frac{\pi}{3} = \sqrt{3}, \quad \tan 2x = \tan \frac{2\pi}{3} = -\sqrt{3}, \quad \tan 3x = \tan \pi = 0
\]
\[
LHS = 0 \cdot (-\sqrt{3}) \cdot \sqrt{3} = 0
\]
\[
RHS = 0 - (-\sqrt{3}) - \sqrt{3} = 0
\]
So, \( x = \frac{\pi}{3} \) is also a solution.
### Conclusion
The solutions we find are \( x = 0 \) and \( x = \frac{\pi}{3} \). To find more solutions, it may involve analyzing the periodicity of the tangent function and solving for increments of \( \pi \).
Thus, the complete solution will be of the form:
\[
x = n\pi \quad \text{for } n \in \mathbb{Z} \quad \text{and} \quad x = \frac{\pi}{3} + k\pi \quad \text{for } k \in \mathbb{Z}.
\]