Start by moving arcsin(y) to the right side:

arcsin(y/2) = 120 - arcsiny

Then take sin of the both sides:

sin(arcsin(y/2)) = sin(120 - arcsiny)

The left side will be y/2. The right side can be simplified using the formula

sin(x-y) = sinxcosy - cosxsiny

sin(120-arcsiny) = sin(120)cos(arcsiny) - cos(120)sin(arcsiny)

Substituting `sin120 = sqrt(3)/2` , `cos120 = -1/2` and `cos(arcsin(y)) = sqrt(1 - y^2)` ,

get

`y/2 = sqrt(3)/2*sqrt(1 - y^2)+1/2*y`

Subtracting y/2 from both sides results in

`sqrt(3)/2 * sqrt(1 - y^2) = 0`

This can only be true if y = 1 or y = -1.

Let's chcek these results by plugging them into the original equation:

arcsin(1) = 90 and arcsin(1/2) = 30. 90 +30 = 120 so y = 1 is a solution of the given equation.

arcsin(-1) = -90 and arcsin(-1/2) = -30. -90 + (-30) is not equal to 120, therefore y = -1 is not a valid solution.

**Answer: y = 1**.