You need to use the formula of cosine of double angle such that:

`cos 4y = cos 2*(2y) = 1 - 2sin^2 (2y)`

Hence, you may write the equation in terms of `sin (2y)` such that:

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

You need to move all terms to one side such that:

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

You should come up with the substitution such that:

`sin (2y)=x`

`2x^2 + x - 1 = 0`

You need to apply quadratic formula such that:

`x_(1,2) = (-1+-sqrt(1+8))/4`

`x_(1,2) = (-1+-3)/4`

`x_(1) =-1`

`x_(2) = 1/2`

You need to find y, hence you need to solve equations:

`sin 2y = -1 =gt 2y = sin^(-1) (-1) =gt 2y = 270^o =gt y = 135^o`

`sin 2y = 1/2 =gt 2y = sin^(-1) (1/2) =gt 2y = 30^o =gt y= 15^o`

`2y = sin^(-1) (1/2) =gt 2y = 180^o - 30^o =gt 2y= 150^o=gt y=75^o`

**Hence, the solutions to equation are `y_1 = 15^o ; y_2 = 75^o ; y_3 = 135^o.` **