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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.

Find the equation of the line tangent to the graph of f(x) = 3 sin(x) - 2 cos(x) at x = -π.

Learn how to find the equation of the tangent line to the graph of f(x) = 3sin(x) - 2cos(x) at x = -π. This problem involves differentiating trigonometric functions to calculate the slope and using the point-slope form to write the equation of the tangent line. Suitable for students in Grades 11-12 or Calculus I.

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