91Ó°ÊÓ

To find the gradient (or the normal vector) of the surface defined by \(F(x, y, z) = xy\sin z - 1 = 0\), compute the partial derivatives of the function \(F\) with respect to \(x\), \(y\), and \(z\): $$\frac{\partial F}{\partial x} = y\sin z$$ $$\frac{\partial F}{\partial y} = x\sin z$$ $$\frac{\partial F}{\partial z} = xy\cos z$$

Now, we need to find the normal vector of the tangent plane at the given points \((1, 2, \pi/6)\) and \((-2, -1, 5\pi/6)\). Evaluate the gradient at these points: $$\nabla F(1, 2, \pi/6) = \langle 2\sin(\pi/6), 1\sin(\pi/6), 2\cos(\pi/6)\rangle = \langle 2, \frac{1}{2}, \sqrt{3}\rangle$$ $$\nabla F(-2, -1, 5\pi/6) = \langle -\sin(5\pi/6), -2\sin(5\pi/6), 2\cos(5\pi/6)\rangle = \langle -\frac{1}{2}, -1, -\sqrt{3}\rangle$$

Using the normal vector and the given points, we can now find the equations of the tangent planes. Recall that the equation of a plane is given by \(A(x-x_0) + B(y-y_0) + C(z-z_0) = 0\), where \((A, B, C)\) is the normal vector, and \((x_0, y_0, z_0)\) is a known point on the plane. For the first point \((1, 2, \pi/6)\) and normal vector \(\langle 2, \frac{1}{2}, \sqrt{3}\rangle\): $$2(x-1) + \frac{1}{2}(y-2) + \sqrt{3}(z-\frac{\pi}{6}) = 0$$ For the second point \((-2, -1, 5\pi/6)\) and normal vector \(\langle -\frac{1}{2}, -1, -\sqrt{3}\rangle\): $$-\frac{1}{2}(x+2) - (y+1) - \sqrt{3}(z-\frac{5\pi}{6}) = 0$$ So, the two tangent plane equations are: $$2(x-1) + \frac{1}{2}(y-2) + \sqrt{3}(z-\frac{\pi}{6}) = 0$$ and $$-\frac{1}{2}(x+2) - (y+1) - \sqrt{3}(z-\frac{5\pi}{6}) = 0$$