The equation sin(y/4)-8cos(y/6)= -3 has to be solved for y.

Substitute y = `12x :`

=> `sin((12x)/4)-8cos((12x)/6)= -3`

=> `sin(3x)-8cos(2x)= -3`

Sin 3x and cos 2x can be written in terms of sin x as follows:

`3*sin x - 4*sin^3x - 8(1-2*sin^2x) = -3`

`3*sin x - 4*sin^3x - 8 +16*sin^2x + 3 = 0`

Substitute z = sin x:

=> `3*z - 4*z^3 - 5 + 16z^2 = 0`

=> `4z^3 - 14z^2-2z^2-10z+7z+5=0`

=> `2z(2z^2-7z-5)-1(2z^2-7z-5)=0`

=> `(2z-1)(2z^2-7z-5)=0`

=> `z = 1/2 and z = 7/4+-sqrt89/4`

Now, z = sin x, but sin x lies in (-1, 1). It is defined for z = 1/2 and `z = 7/4-sqrt89/4`

`sin x = 1/2 and sinx = 7/4-sqrt89/4`

=> `x = sin^-1(1/2) and x = sin^-1(7/4-sqrt 89/4)`

As y = 12x ,

`y = 12*sin^-1(1/2) and y = 12sin^-1(7/4-sqrt 89/4)`

The solution of the required equation is `y = 12*sin^-1(1/2)` and `y = 12sin^-1(7/4-sqrt 89/4).`