The problem asks us to find the equation of the tangent plane to the given surface $z = 2e^{3y} \cos 2x$ at the point $(\pi/3, 0, -1)$.

1. Problem Description

The problem asks us to find the equation of the tangent plane to the given surface $z = 2e^{3y} \cos 2x$ at the point $(\pi/3, 0, -1)$.

2. Solution Steps

Let $F(x, y, z) = 2e^{3y} \cos 2x - z$.

Then, the surface is given by $F(x, y, z) = 0$.

We need to find the partial derivatives of $F$ with respect to $x$, $y$, and $z$.

$F_x = \frac{\partial F}{\partial x} = 2e^{3y} (-2 \sin 2x) = -4e^{3y} \sin 2x$

$F_y = \frac{\partial F}{\partial y} = 2 (3 e^{3y}) \cos 2x = 6e^{3y} \cos 2x$

$F_z = \frac{\partial F}{\partial z} = -1$

Now, we evaluate the partial derivatives at the point $(\pi/3, 0, -1)$:

$F_x (\pi/3, 0, -1) = -4 e^{3(0)} \sin (2\pi/3) = -4 (1) (\sqrt{3}/2) = -2\sqrt{3}$

$F_y (\pi/3, 0, -1) = 6 e^{3(0)} \cos (2\pi/3) = 6 (1) (-1/2) = -3$

$F_z (\pi/3, 0, -1) = -1$

The equation of the tangent plane at $(\pi/3, 0, -1)$ is given by:

$F_x (x - x_0) + F_y (y - y_0) + F_z (z - z_0) = 0$

where $(x_0, y_0, z_0) = (\pi/3, 0, -1)$.

So, we have:

$-2\sqrt{3}(x - \pi/3) - 3(y - 0) - 1(z - (-1)) = 0$

$-2\sqrt{3}x + 2\sqrt{3}\pi/3 - 3y - z - 1 = 0$

$-2\sqrt{3}x - 3y - z + (2\sqrt{3}\pi/3 - 1) = 0$

$2\sqrt{3}x + 3y + z = \frac{2\sqrt{3}\pi}{3} - 1$

3. Final Answer

$2\sqrt{3}x + 3y + z = \frac{2\sqrt{3}\pi}{3} - 1$

or

$2\sqrt{3}x + 3y + z - (\frac{2\pi \sqrt{3}}{3} - 1) = 0$