91Ó°ÊÓ

To find the points with horizontal tangent planes, we need to find the partial derivatives of z with respect to x and y: $$\frac{\partial{z}}{\partial{x}} = -2 \sin{(2x)} \sin{(y)}$$ $$\frac{\partial{z}}{\partial{y}} = \cos{(2x)} \cos{(y)}$$

We now need to solve the following equations to find points with horizontal tangent planes: $$-2 \sin{(2x)} \sin{(y)} = 0$$ $$\cos{(2x)} \cos{(y)} = 0$$

Analyzing the equations, we have two cases for each equation: For the first equation: 1. \(\sin{(2x)} = 0\) 2. \(\sin{(y)} = 0\) For the second equation: 1. \(\cos{(2x)} = 0\) 2. \(\cos{(y)} = 0\) We now solve each case: 1. For \(\sin{(2x)} = 0\), \(2x = k\pi\), where k is an integer. So, \(x=\frac{k\pi}{2}\). 2. For \(\sin{(y)} = 0\), \(y = l\pi\), where l is an integer. 3. For \(\cos{(2x)} = 0\), \(2x = (2m+1)\frac{\pi}{2}\), where m is an integer. So, \(x = (2m+1)\frac{\pi}{4}\). 4. For \(\cos{(y)} = 0\), \(y = (2n+1)\frac{\pi}{2}\), where n is an integer.

Using above solutions, we apply the given region for x and y: \(-\pi \leq x \leq \pi,-\pi \leq y \leq \pi\) 1. For x, when k = -2, -1, 0, 1, 2, we have: \(x = -\pi, -\frac{\pi}{2}, 0, \frac{\pi}{2}, \pi\) 2. For y, when l = -1, 0, 1, we have: \(y = -\pi, 0, \pi\) 3. For x, when m = -1, 0, 1, we have: \(x = -\frac{3\pi}{4}, -\frac{\pi}{4}, \frac{\pi}{4}, \frac{3\pi}{4}\) 4. For y, when n = -1, 0, 1, we have: \(y = -\frac{3\pi}{2}, -\frac{\pi}{2}, \frac{\pi}{2}, \frac{3\pi}{2}\)

Combine the possible x and y values to find points (x, y) with horizontal tangent planes on the given surface: From partial derivative with respect to x: \((x, y) = \left(-\pi, -\pi\right), \left(-\pi, 0\right), \left(-\pi, \pi\right), \left(-\frac{\pi}{2}, -\pi\right), \left(-\frac{\pi}{2}, 0\right), \left(-\frac{\pi}{2}, \pi\right),\) \(\left(0, -\pi\right), \left(0, 0\right), \left(0, \pi\right), \left(\frac{\pi}{2}, -\pi\right), \left(\frac{\pi}{2}, 0\right), \left(\frac{\pi}{2}, \pi\right),\) \(\left(\pi, -\pi\right), \left(\pi, 0\right), \left(\pi, \pi\right)\) From partial derivative with respect to y: \((x, y) = \left(-\frac{3\pi}{4}, -\frac{3\pi}{2}\right), \left(-\frac{3\pi}{4}, -\frac{\pi}{2}\right),\) \(\left(-\frac{3\pi}{4}, \frac{\pi}{2}\right), \left(-\frac{3\pi}{4}, \frac{3\pi}{2}\right),\) \(\left(-\frac{\pi}{4}, -\frac{3\pi}{2}\right), \left(-\frac{\pi}{4}, -\frac{\pi}{2}\right),\) \(\left(-\frac{\pi}{4}, \frac{\pi}{2}\right), \left(-\frac{\pi}{4}, \frac{3\pi}{2}\right),\) \(\left(\frac{\pi}{4}, -\frac{3\pi}{2}\right), \left(\frac{\pi}{4}, -\frac{\pi}{2}\right),\) \(\left(\frac{\pi}{4}, \frac{\pi}{2}\right), \left(\frac{\pi}{4}, \frac{3\pi}{2}\right),\) \(\left(\frac{3\pi}{4}, -\frac{3\pi}{2}\right), \left(\frac{3\pi}{4}, -\frac{\pi}{2}\right),\) \(\left(\frac{3\pi}{4}, \frac{\pi}{2}\right), \left(\frac{3\pi}{4}, \frac{3\pi}{2}\right)\) However, some of the y values in this case exceed the given region of \(-\pi \leq y \leq \pi\). After removing such points, we have: \((x, y) = \left(-\frac{3\pi}{4}, -\frac{\pi}{2}\right), \left(-\frac{3\pi}{4}, \frac{\pi}{2}\right),\) \(\left(-\frac{\pi}{4}, -\frac{\pi}{2}\right), \left(-\frac{\pi}{4}, \frac{\pi}{2}\right),\) \(\left(\frac{\pi}{4}, -\frac{\pi}{2}\right), \left(\frac{\pi}{4}, \frac{\pi}{2}\right),\) \(\left(\frac{3\pi}{4}, -\frac{\pi}{2}\right), \left(\frac{3\pi}{4}, \frac{\pi}{2}\right)\) These are the points at which the given surfaces have horizontal tangent planes in the given region.