To solve the equation cos(2x)cos(5x) = cos(3x)cos(4x), we can use the trigonometric identity:

cos(A)cos(B) = 1/2[cos(A + B) + cos(A - B)]

Applying this identity to the left side of the equation, we have:

cos(2x)cos(5x) = 1/2[cos(2x + 5x) + cos(2x - 5x)]
= 1/2[cos(7x) + cos(-3x)]
= 1/2[cos(7x) + cos(3x)]

Now, let's apply the same identity to the right side of the equation:

cos(3x)cos(4x) = 1/2[cos(3x + 4x) + cos(3x - 4x)]
= 1/2[cos(7x) + cos(-x)]
= 1/2[cos(7x) + cos(x)]

Now, we have:

1/2[cos(7x) + cos(3x)] = 1/2[cos(7x) + cos(x)]

Since the left side and the right side are equal, we can cancel out the 1/2 factor:

cos(7x) + cos(3x) = cos(7x) + cos(x)

Now, we can subtract cos(7x) from both sides:

cos(3x) = cos(x)

To solve this equation, we can use the identity:

cos(A) = cos(B) if and only if A = B + 2πn or A = -B + 2πn, where n is an integer.

Applying this identity to our equation, we have:

3x = x + 2πn or 3x = -x + 2πn

Simplifying each equation, we get:

2x = 2πn or 4x = 2πn

Dividing both sides by 2, we have:

x = πn or 2x = πn

Simplifying further, we get:

x = πn or x = πn/2

Therefore, the solutions to the equation cos(2x)cos(5x) = cos(3x)cos(4x) are:

x = πn or x = πn/2, where n is an integer.