To find the equation of the tangent line to the curve $y = 6 \sin x + 7 \sec x$ at $x = \pi$, we will proceed as follows:

1. Find the derivative of the function:

The derivative $\frac{dy}{dx}$ will give us the slope of the tangent line.

• Derivative of $6 \sin x$ is $6 \cos x$.
• Derivative of $7 \sec x$ is $7 \sec x \tan x$.

Thus, the derivative is:

$\frac{dy}{dx} = 6 \cos x + 7 \sec x \tan x$

2. Evaluate the derivative at $x = \pi$:

Now, substitute $x = \pi$ into the derivative to find the slope at that point.

• $\cos(\pi) = -1$
• $\sec(\pi) = -1$ (since $\sec(x) = \frac{1}{\cos(x)}$)
• $\tan(\pi) = 0$

Therefore:

$\frac{dy}{dx}\bigg|_{x = \pi} = 6(-1) + 7(-1)(0) = -6$

So, the slope of the tangent line at $x = \pi$ is $m = -6$.

3. Find the point on the curve at $x = \pi$:

To find the equation of the tangent line, we need the point on the curve at $x = \pi$. Substitute $x = \pi$ into the original function:

$y = 6 \sin(\pi) + 7 \sec(\pi)$

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

$y = 6(0) + 7(-1) = -7$

Thus, the point is $(\pi, -7)$.

4. Equation of the tangent line:

The equation of the tangent line is given by the point-slope form:

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

Substitute $m = -6$, $x_1 = \pi$, and $y_1 = -7$:

$y + 7 = -6(x - \pi)$

Simplify:

$y = -6x + 6\pi - 7$

Thus, the equation of the tangent line is:

$y = -6x + 6\pi - 7$

Would you like further details on this solution, or do you have any questions? Here are five related questions you might find interesting:

1. How do you find the derivative of trigonometric functions like $\sec(x)$?
2. What is the general form of a tangent line equation?
3. How does the derivative represent the slope of a function at a given point?
4. Can you apply this method to other points on the curve?
5. How do you handle more complex trigonometric derivatives?

Tip: Always evaluate trigonometric functions like sine, cosine, and secant carefully at special angles like $\pi$, $\frac{\pi}{2}$, etc., as their values can simplify the calculations significantly.