To find the equation of the tangent to the curve at the given point, let's go through the following steps.

Problem Recap

We have: $x = \cos \theta + \sin 2\theta + 2, \quad y = \sin \theta + \cos 2\theta + 1$ with $\theta = \pi$.

Steps to Solve

1. Calculate $x$ and $y$ at $\theta = \pi$:

   Substitute $\theta = \pi$ into each equation to find the coordinates $(x, y)$ at this point.

   • $x = \cos(\pi) + \sin(2 \cdot \pi) + 2$ $= -1 + 0 + 2 = 1$

   • $y = \sin(\pi) + \cos(2 \cdot \pi) + 1$ $= 0 + 1 + 1 = 2$

   So, the point on the curve at $\theta = \pi$ is $(1, 2)$.

2. Differentiate $x$ and $y$ with respect to $\theta$:

   To find the slope of the tangent, we need $\frac{dy}{d\theta}$ and $\frac{dx}{d\theta}$, then compute $\frac{dy}{dx}$ as $\frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}$.

   • Differentiate $x$ with respect to $\theta$: $\frac{dx}{d\theta} = -\sin \theta + 2 \cos 2\theta$

   • Differentiate $y$ with respect to $\theta$: $\frac{dy}{d\theta} = \cos \theta - 2 \sin 2\theta$

3. Evaluate $\frac{dx}{d\theta}$ and $\frac{dy}{d\theta}$ at $\theta = \pi$:

   • $\frac{dx}{d\theta}$ at $\theta = \pi$: $= -\sin(\pi) + 2 \cos(2\pi) = 0 + 2 \cdot 1 = 2$

   • $\frac{dy}{d\theta}$ at $\theta = \pi$: $= \cos(\pi) - 2 \sin(2\pi) = -1 - 0 = -1$

4. Find $\frac{dy}{dx}$ at $\theta = \pi$:

   $\frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} = \frac{-1}{2} = -\frac{1}{2}$

   So, the slope of the tangent line at the point $(1, 2)$ is $-\frac{1}{2}$.

5. Write the equation of the tangent line:

   Using the point-slope form of a line $y - y_1 = m(x - x_1)$, with $m = -\frac{1}{2}$ and point $(1, 2)$: $y - 2 = -\frac{1}{2}(x - 1)$

   Simplifying: $y = -\frac{1}{2}x + \frac{1}{2} + 2$ $y = -\frac{1}{2}x + \frac{5}{2}$

Final Answer

The equation of the tangent line to the curve at $\theta = \pi$ is: $y = -\frac{1}{2}x + \frac{5}{2}$

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

1. How do we derive the derivatives of trigonometric functions in parametric equations?
2. What is the process to differentiate parametric equations in terms of a parameter?
3. How would the equation change if we had a different value for $\theta$?
4. Why is the point-slope form useful for finding tangent lines?
5. Can we find the normal line at this point instead of the tangent?

Tip: When differentiating trigonometric functions, make sure to apply the chain rule properly if there are coefficients within the argument, such as $2\theta$.