91Ó°ÊÓ

The given function is \(z = \cos(2x)\sin(y)\). Let's compute the partial derivatives with respect to x and y: 1. Partial derivative with respect to x: $$\frac{\partial z}{\partial x} = -2\sin(2x)\sin(y)$$ 2. Partial derivative with respect to y: $$\frac{\partial z}{\partial y} = \cos(2x)\cos(y)$$

Horizontal tangent plane exists at points where both partial derivatives are equal to zero: $$ \begin{cases} -2\sin(2x)\sin(y) = 0 \\ \cos(2x)\cos(y) = 0 \end{cases} $$

We'll solve the system of equations by analyzing each equation separately and finding the points where the equations are simultaneously satisfied. 1. Analyze the first equation: $$-2\sin(2x)\sin(y) = 0$$ This equation is true when either \(\sin(2x) = 0\) or \(\sin(y)=0\). a) When \(\sin(2x) = 0\): $$2x = n\pi, \quad x = \frac{n\pi}{2}, \quad n \in \mathbb{Z}$$ b) When \(\sin(y) = 0\): $$y = m\pi, \quad m \in \mathbb{Z}$$ 2. Analyze the second equation: $$\cos(2x)\cos(y) = 0$$ This equation is true when either \(\cos(2x) = 0\) or \(\cos(y)=0\). a) When \(\cos(2x) = 0\): $$2x = (2n+1)\frac{\pi}{2}, \quad x = (2n+1)\frac{\pi}{4}, \quad n \in \mathbb{Z}$$ b) When \(\cos(y) = 0\): $$y = (2m+1)\frac{\pi}{2}, \quad m \in \mathbb{Z}$$ Now we need to find the points where both above conditions (1 and 2) are satisfied in the region \(-\pi \leq x \leq \pi,-\pi \leq y \leq \pi\).

We can see that for every point \((x,y)\) such that: $$ x = \frac{n\pi}{2}, y = m\pi $$ or $$ x = (2n+1)\frac{\pi}{4}, y = (2m+1)\frac{\pi}{2} $$ we have a horizontal tangent plane on the surface. Since we have a restricted region \(-\pi \leq x \leq \pi,-\pi \leq y\leq \pi\), we can find the points in this region: Points with horizontal tangent planes are: $$ (-\pi, -\pi) , (-\pi,0), (-\pi, \pi), \left(-\frac{\pi}{2}, -\frac{\pi}{2}\right), \left(-\frac{\pi}{2}, \frac{\pi}{2}\right), (0, -\pi), (0,0), (0, \pi), \left(\frac{\pi}{2}, -\frac{\pi}{2}\right), \left(\frac{\pi}{2}, \frac{\pi}{2}\right), (\pi, -\pi) , (\pi,0), (\pi, \pi) $$