The problem asks to find the equation of the line tangent to the graph of the function $f(x) = 3 \sin(x) - 2 \cos(x)$ at $x = -\pi$.

Steps to solve:

1. Find the derivative of the function $f(x)$. This will give us the slope of the tangent line at any point $x$.

   $f'(x) = \frac{d}{dx} [3 \sin(x) - 2 \cos(x)] = 3 \cos(x) + 2 \sin(x)$

2. Evaluate the derivative at $x = -\pi$ to find the slope of the tangent line at this point.

   $f'(-\pi) = 3 \cos(-\pi) + 2 \sin(-\pi)$

   Since $\cos(-\pi) = -1$ and $\sin(-\pi) = 0$, we get:

   $f'(-\pi) = 3(-1) + 2(0) = -3$

   So, the slope of the tangent line is $m = -3$.

3. Find the value of the function at $x = -\pi$ to determine the point of tangency. This gives us the $y$-coordinate of the point on the curve.

   $f(-\pi) = 3 \sin(-\pi) - 2 \cos(-\pi)$

   Since $\sin(-\pi) = 0$ and $\cos(-\pi) = -1$, we get:

   $f(-\pi) = 3(0) - 2(-1) = 2$

   The point of tangency is $(-\pi, 2)$.

4. Write the equation of the tangent line using the point-slope form:

   $y - y_1 = m(x - x_1)$

   Substituting $m = -3$, $x_1 = -\pi$, and $y_1 = 2$, we get:

   $y - 2 = -3(x + \pi)$

   Simplifying:

   $y = -3(x + \pi) + 2$ $y = -3x - 3\pi + 2$

So, the equation of the tangent line is: $y = -3x - 3\pi + 2$

Would you like more details on any step? Here are some related questions you might explore:

1. How is the derivative used to find the slope of a tangent line?
2. What is the point-slope form of a line equation?
3. Why do trigonometric functions have specific values at angles like $\pi$ and $-\pi$?
4. How would the tangent line differ if the slope at $x = -\pi$ were positive?
5. How does the chain rule apply when differentiating trigonometric functions?

Tip: Remember, the derivative tells you how fast the function is changing at a given point, which is why it helps define the slope of the tangent line.