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We are tasked with finding the equation of the tangent line to the function \( f(x) = -3 \sec(x) \) at \( x = \frac{11\pi}{6} \).

### Steps to solve:

1. **Find the value of \( f(x) \) at \( x = \frac{11\pi}{6} \):**

   \[
   f(x) = -3 \sec(x)
   \]
   Substituting \( x = \frac{11\pi}{6} \):
   
   \[
   f\left( \frac{11\pi}{6} \right) = -3 \sec \left( \frac{11\pi}{6} \right)
   \]
   Now, \( \sec(x) = \frac{1}{\cos(x)} \), and \( \cos\left( \frac{11\pi}{6} \right) = \cos\left( \frac{-\pi}{6} \right) = \frac{\sqrt{3}}{2} \).
   
   Therefore:
   
   \[
   \sec\left( \frac{11\pi}{6} \right) = \frac{1}{\frac{\sqrt{3}}{2}} = \frac{2}{\sqrt{3}}
   \]
   So:
   
   \[
   f\left( \frac{11\pi}{6} \right) = -3 \times \frac{2}{\sqrt{3}} = -\frac{6}{\sqrt{3}} = -2\sqrt{3}
   \]

2. **Find the derivative \( f'(x) \):**

   To find the slope of the tangent line, we need to differentiate \( f(x) = -3 \sec(x) \). The derivative of \( \sec(x) \) is \( \sec(x) \tan(x) \), so:

   \[
   f'(x) = -3 \sec(x) \tan(x)
   \]
   Now, substitute \( x = \frac{11\pi}{6} \) into \( f'(x) \):

   \[
   f'\left( \frac{11\pi}{6} \right) = -3 \sec\left( \frac{11\pi}{6} \right) \tan\left( \frac{11\pi}{6} \right)
   \]
   We already know \( \sec\left( \frac{11\pi}{6} \right) = \frac{2}{\sqrt{3}} \). Next, \( \tan\left( \frac{11\pi}{6} \right) = \tan\left( \frac{-\pi}{6} \right) = -\frac{1}{\sqrt{3}} \).
   
   Therefore:
   
   \[
   f'\left( \frac{11\pi}{6} \right) = -3 \times \frac{2}{\sqrt{3}} \times \left( -\frac{1}{\sqrt{3}} \right) = -3 \times \frac{-2}{3} = 2
   \]

3. **Write the equation of the tangent line:**

   The equation of the tangent line is given by:

   \[
   y - f(x_0) = f'(x_0)(x - x_0)
   \]
   Here, \( x_0 = \frac{11\pi}{6} \), \( f(x_0) = -2\sqrt{3} \), and \( f'(x_0) = 2 \). Substituting these values:

   \[
   y + 2\sqrt{3} = 2\left( x - \frac{11\pi}{6} \right)
   \]
   Simplifying:

   \[
   y = 2x - \frac{11\pi}{3} - 2\sqrt{3}
   \]

Thus, the equation of the tangent line is:

\[
y = 2x - \frac{11\pi}{3} - 2\sqrt{3}
\]

Would you like any further details or clarifications?

---

### Follow-up questions:

1. How do you differentiate trigonometric functions like \( \sec(x) \) and \( \tan(x) \)?
2. What is the geometric interpretation of a tangent line?
3. Can you explain how to find \( \sec(x) \) and \( \tan(x) \) using a unit circle?
4. What are the common steps to find the equation of a tangent line for any function?
5. How would the solution change if the given function were different, such as \( f(x) = -3 \cos(x) \)?

**Tip:** When working with trigonometric functions, always check if the given angle is within standard unit circle values to make calculations faster!

Given f(x) = -3 sec(x), write the equation of the line tangent to y = f(x) when x = 11π/6.

Find the equation of the tangent line to the function f(x) = -3 sec(x) at x = 11π/6. This problem covers differentiation, secant and tangent trigonometric functions, and applying the tangent line equation. Ideal for Grades 11-12 calculus students.

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