# solve sin2y=cos4y for y, where 0<y<360. hint: cos 4y=cos(2y+2y)

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You need to move all terms to one side and then you should add and subtract the constant terms 1 such that:

`sin 2y - cos 4y + 1 - 1 = 0`

You need to remember the formula of half of angle such that `1 - cos 4y = 2 sin^2 2y` :

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

You should come up with the substitution such that:

`sin 2y = x`

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

You ned to use quadratic formula such that:

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

`x_1 = (-1+3)/4 =gt x_1 = 1/2`

`x_2 = (-1-3)/4 =gt x_2 = -1`

You need to solve for y, in interval`(0^o,360^o),` the equations `sin 2y = 1/2 ` and sin `2y = -1 ` such that:

`sin 2y = 1/2`

You need to remember that the values of the sine function are positive in quadrants 1 and 2 such that:

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

and `2y = 180^o - 30^o =gt 2y = 15o^o =gt y = 75^o`

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

**Hence, evaluating the solutions to given equation over interval `(0^o,360^o)` yields `y = 15^o`, `y = 75^o` and `y = 135^o` .**