To solve the problem and find the points $(x, y)$ where the curve $x(t) = 2\cos(t)$ and $y(t) = 2\sin(2t)$ has a horizontal tangent, we follow these steps:

Step 1: Tangent Line Condition

A horizontal tangent occurs when the derivative of $y(t)$ with respect to $t$ is zero. That is: $\frac{dy}{dt} = 0$

Step 2: Derivatives of $y(t)$

The given $y(t) = 2\sin(2t)$. Compute its derivative: $\frac{dy}{dt} = 2 \cdot \cos(2t) \cdot 2 = 4\cos(2t)$

Set $\frac{dy}{dt} = 0$: $4\cos(2t) = 0 \implies \cos(2t) = 0$

Step 3: Solve for $t$

$\cos(2t) = 0$ when: $2t = \frac{\pi}{2} + n\pi, \, n \in \mathbb{Z}$ $t = \frac{\pi}{4} + \frac{n\pi}{2}, \, n \in \mathbb{Z}$

Step 4: Compute $(x, y)$

For each $t$, compute $x(t)$ and $y(t)$: $x(t) = 2\cos(t), \quad y(t) = 2\sin(2t)$

• For $t = \frac{\pi}{4}$: $x\left(\frac{\pi}{4}\right) = 2\cos\left(\frac{\pi}{4}\right) = 2 \cdot \frac{\sqrt{2}}{2} = \sqrt{2}$ $y\left(\frac{\pi}{4}\right) = 2\sin\left(2 \cdot \frac{\pi}{4}\right) = 2\sin\left(\frac{\pi}{2}\right) = 2$

• For $t = \frac{3\pi}{4}$: $x\left(\frac{3\pi}{4}\right) = 2\cos\left(\frac{3\pi}{4}\right) = 2 \cdot \left(-\frac{\sqrt{2}}{2}\right) = -\sqrt{2}$ $y\left(\frac{3\pi}{4}\right) = 2\sin\left(2 \cdot \frac{3\pi}{4}\right) = 2\sin\left(\frac{3\pi}{2}\right) = -2$

Final Points

The points where the curve has a horizontal tangent are: $(\sqrt{2}, 2) \quad \text{and} \quad (-\sqrt{2}, -2)$

Let me know if you'd like a deeper explanation or further assistance. Here are some related questions for you to explore:

1. What is the significance of a horizontal tangent in a parametric curve?
2. How would you find the vertical tangents for this curve?
3. What happens to the curve’s behavior at these points?
4. Can you determine the symmetry of the curve?
5. How would adding a constant to $y(t)$ affect the horizontal tangents?

Tip: Always start by finding critical points ($\frac{dy}{dt} = 0$) to analyze tangent properties in parametric equations.