The problem in the image asks for the equations of the tangent lines to the curve $y = \cos(x)$ at several points $x = -\frac{\pi}{2}, 0, \text{and} \frac{\pi}{2}$, and to graph the curve over a specific interval along with its tangent lines.

Let's focus on part of the problem that asks:

"What is the equation of the tangent line to the curve at $x = -\frac{\pi}{2}$?"

Steps to solve:

1. Find the derivative of the function $y = \cos(x)$. This gives the slope of the tangent line. $\frac{dy}{dx} = -\sin(x)$

2. Evaluate the derivative at $x = -\frac{\pi}{2}$ to get the slope of the tangent line: $\frac{dy}{dx}\bigg|_{x=-\frac{\pi}{2}} = -\sin\left(-\frac{\pi}{2}\right) = -(-1) = 1$ So, the slope of the tangent line is $1$.

3. Find the y-coordinate at $x = -\frac{\pi}{2}$: $y = \cos\left(-\frac{\pi}{2}\right) = 0$ So, the point of tangency is $\left(-\frac{\pi}{2}, 0\right)$.

4. Use the point-slope form of the equation of a line: $y - y_1 = m(x - x_1)$ where $m = 1$ is the slope, and $\left(x_1, y_1\right) = \left(-\frac{\pi}{2}, 0\right)$ is the point of tangency. Substituting the values: $y - 0 = 1\left(x - \left(-\frac{\pi}{2}\right)\right)$ $y = x + \frac{\pi}{2}$

Thus, the equation of the tangent line at $x = -\frac{\pi}{2}$ is: $y = x + \frac{\pi}{2}$

Would you like more details on graphing the curve and the tangent lines or explanations for the other points?

Here are some related questions for deeper understanding:

1. How do we determine the slope of a tangent line from the derivative?
2. What is the significance of evaluating the function at a particular point?
3. How do you find the equation of a tangent line if the function is more complex?
4. Can the process be generalized for other trigonometric functions like $y = \sin(x)$?
5. What are the possible uses of tangent lines in applied problems?

Tip: The point-slope form of a line is very useful for quickly finding the equation of a tangent line once the slope is known.