Problem: Find the equation of tangents to the curve y = cos(xy), -2π ≤ x ≤ 2π that are parallel to the line x + 2y = 0.

Approach:
To find the equation of the tangents that are parallel to a given line, we need to follow these steps:
1. Find the slope of the given line.
2. Find the derivative of the curve with respect to x.
3. Set the derivative equal to the slope of the given line.
4. Solve the resulting equation to find the x-values where the tangents are parallel.
5. Substitute the x-values into the original curve equation to find the corresponding y-values.
6. Use the point-slope form of a linear equation to find the equation of the tangents.

Step 1: Finding the slope of the given line
The given line is x + 2y = 0, which can be rewritten in slope-intercept form as y = -1/2x. The slope of this line is -1/2.

Step 2: Finding the derivative of the curve
Given curve: y = cos(xy)
To find the derivative, we need to use the chain rule.
Let u = xy
Then, y = cos(u)
Using the chain rule, dy/dx = dy/du * du/dx
dy/du = -sin(u)
du/dx = y + x(dy/dx) [using the product rule]
Substituting the values, we get:
dy/dx = -sin(xy) * (y + x(dy/dx))
dy/dx + sin(xy) * x(dy/dx) = -sin(xy) * y
dy/dx(1 + sin(xy) * x) = -sin(xy) * y
dy/dx = -sin(xy) * y / (1 + sin(xy) * x)

Step 3: Setting the derivative equal to the slope of the given line
Since the tangents are parallel to the line x + 2y = 0, the slope of the tangents will also be -1/2.
Therefore, we set the derivative equal to -1/2:
-sin(xy) * y / (1 + sin(xy) * x) = -1/2

Step 4: Solving the equation to find x-values
To solve the equation -sin(xy) * y / (1 + sin(xy) * x) = -1/2 for x, we can cross-multiply:
-2 * sin(xy) * y = 1 + sin(xy) * x
-2sin(xy)y - sin(xy)x = 1
x(-sin(xy)) - 2y(sin(xy)) = 1
x(-sin(xy)) = 2y(sin(xy)) + 1

Now, we can solve this equation to find the x-values where the tangents are parallel.

Step 5: Finding the corresponding y-values
After finding the x-values, we substitute them into the original curve equation y = cos(xy) to find the corresponding y-values.

Step 6: Finding the equation of the tangents
Once we have the x and y-values, we can use the point-slope form of a linear equation