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We are given the curve \( y = f(x) \) defined parametrically by the equations:
\[
x(t) = \sin(2t) + \cos(2t), \quad y(t) = 3t^2 - 1, \quad t \in [0, 2\pi].
\]
We need to determine the point on the curve where the tangent is parallel to the \( x \)-axis. Additionally, we are tasked with finding the equations of the tangent and normal lines at that point.

### Steps to Solve:

1. **Find the derivative \( \frac{dy}{dx} \)**:
   We first calculate the parametric derivatives:
   \[
   \frac{dx}{dt} = \frac{d}{dt}[\sin(2t) + \cos(2t)] = 2\cos(2t) - 2\sin(2t),
   \]
   \[
   \frac{dy}{dt} = \frac{d}{dt}[3t^2 - 1] = 6t.
   \]
   The slope of the tangent line is given by:
   \[
   \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{6t}{2\cos(2t) - 2\sin(2t)} = \frac{3t}{\cos(2t) - \sin(2t)}.
   \]
   To find where the tangent is parallel to the \( x \)-axis, we set \( \frac{dy}{dx} = 0 \). This implies that:
   \[
   3t = 0 \quad \Rightarrow \quad t = 0.
   \]

2. **Find the point on the curve when \( t = 0 \)**:
   We substitute \( t = 0 \) into the parametric equations for \( x(t) \) and \( y(t) \):
   \[
   x(0) = \sin(0) + \cos(0) = 1, \quad y(0) = 3(0)^2 - 1 = -1.
   \]
   Therefore, the point on the curve is \( (1, -1) \).

3. **Equation of the tangent line**:
   Since the tangent is parallel to the \( x \)-axis, its slope is 0. Therefore, the equation of the tangent line is:
   \[
   y = -1.
   \]

4. **Equation of the normal line**:
   The normal line is perpendicular to the tangent, which means its slope is undefined (a vertical line). Hence, the equation of the normal line is:
   \[
   x = 1.
   \]

### Final Answer:
- The point where the tangent is parallel to the \( x \)-axis is \( (1, -1) \).
- The equation of the tangent line is \( y = -1 \).
- The equation of the normal line is \( x = 1 \).

Would you like more details or have any other questions?

Here are some related questions to further explore the topic:
1. How do we interpret the geometric meaning of the normal line to a curve?
2. What if we had a different range for \( t \)? How would the solution change?
3. How can we find the points where the tangent is parallel to the \( y \)-axis?
4. How would the solution change if the parametric equations were more complex?
5. How does this approach extend to curves in three dimensions?

**Tip**: Always check the domain of the parameter to ensure that the points you're finding lie within the given interval.

Дадена е кривата y = f(x) со параметарските равенки: x(t) = sin 2t + cos 2t, y(t) = 3t^2 - 1, t ∈ [0, 2π]. Определете точка од кривата во која тангентата на кривата е паралелна со x-оската. Напишете ги равенките на тангентата и нормалата на кривата во добиената точка.

Explore how to determine the point on a parametric curve where the tangent line is parallel to the x-axis, and find the equations of both the tangent and the normal lines at that point. This problem involves parametric equations, derivatives, and concepts of tangent and normal lines, suitable for university-level calculus.

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