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.