To solve this equation, we first need to use the double angle identity for sine and cosine:

sin(2y) = 2sin(y)cos(y)
cos(2y) = cos^2(y) - sin^2(y) = 1 - 2sin^2(y)

Now, we substitute these identities into the equation:

2(2sin(y)cos(y)) - 2(1 - 2sin^2(y))cos(y) = cos(y)
4sin(y)cos(y) - 2cos(y) + 4sin^2(y)cos(y) = cos(y)
4sin(y)cos(y) + 4sin^2(y)cos(y) - 2cos(y) = cos(y)
4sin(y)cos(y)(1 + 2sin(y)) - 2cos(y) = cos(y)

Now we can factor out a common term:

2cos(y)(2sin(y)(1 + 2sin(y)) - 1) = cos(y)

Dividing both sides by cos(y), we get:

2sin(y)(1 + 2sin(y)) - 1 = 1
2sin(y) + 4sin^2(y) - 1 = 1
4sin^2(y) + 2sin(y) - 2 = 0

Now we have a quadratic equation in terms of sin(y). We can solve this equation using the quadratic formula:

sin(y) = (-b ± √(b^2 - 4ac)) / 2a
sin(y) = (-2 ± √(2^2 - 44(-2))) / (2*4)
sin(y) = (-2 ± √(4 + 32)) / 8
sin(y) = (-2 ± √36) / 8
sin(y) = (-2 ± 6) / 8

This gives us two possible solutions for sin(y):

1) sin(y) = (6 - 2) / 8 = 4 / 8 = 1/2
y = arcsin(1/2) = π/6

2) sin(y) = (-6 - 2) / 8 = -8 / 8 = -1
y = arcsin(-1) = -π/2

Therefore, the solutions to the equation are y = π/6 and y = -π/2.