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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 \(2\theta\).

Find an equation of the tangent to the curve at the point corresponding to the given value of the parameter.

x = cos(θ) + sin(2θ) + 2, y = sin(θ) + cos(2θ) + 1, θ = π

Learn how to find the equation of the tangent line to a parametric curve defined by x = cos(θ) + sin(2θ) + 2 and y = sin(θ) + cos(2θ) + 1 at θ = π. This step-by-step solution covers essential calculus concepts, including derivatives of trigonometric functions and the use of the point-slope formula for tangent lines.

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