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The problem provided is:

Solve the equation for \(-\pi \leq \theta \leq \pi\):
\[
3 \tan \theta - 3 = 5 \tan \theta - 1
\]

### Step-by-Step Solution:

1. **Isolate the \(\tan \theta\) terms:**

   Start by subtracting \(5 \tan \theta\) from both sides:
   \[
   3 \tan \theta - 5 \tan \theta - 3 = -1
   \]
   Simplifying:
   \[
   -2 \tan \theta - 3 = -1
   \]

2. **Solve for \(\tan \theta\):**

   Add 3 to both sides:
   \[
   -2 \tan \theta = 2
   \]
   Now, divide both sides by \(-2\):
   \[
   \tan \theta = -1
   \]

3. **Find the values of \(\theta\):**

   The equation \(\tan \theta = -1\) has solutions where the tangent is \(-1\). Recall that the tangent function is periodic with a period of \(\pi\), and the principal solutions occur at:
   \[
   \theta = \frac{-3\pi}{4}, \frac{\pi}{4}
   \]
   These solutions are within the interval \(-\pi \leq \theta \leq \pi\).

### Final Answer:
\[
\theta = -\frac{3\pi}{4}, \frac{\pi}{4}
\]

Would you like further details or have any questions?

Here are some related questions to consider:
1. How does the periodicity of the tangent function affect its solutions?
2. What is the general solution for \(\tan \theta = -1\)?
3. Why do we only consider solutions within the interval \([- \pi, \pi]\)?
4. How would the solution differ if the equation was \(\tan \theta = 1\)?
5. Can you extend this method to solve more complex trigonometric equations?

**Tip**: When solving trigonometric equations, always be mindful of the interval in which the solution is required, as trigonometric functions are periodic.

Solve the equation for −π ≤ θ ≤ π: 3 tan θ - 3 = 5 tan θ - 1

Learn how to solve the trigonometric equation 3 tan(θ) - 3 = 5 tan(θ) - 1 within the interval −π ≤ θ ≤ π. The step-by-step solution involves isolating tan(θ), solving for θ, and finding principal solutions for the tangent function. Suitable for high school students studying trigonometry.

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