Help... Sin 2y = Cos 4y solve for y...?

Question

Help I need to solve Sin 2y = Cos 4Y for y where 0 < y < 360 Unfortunatly I have no idea where to start... help!

Answer 1

sin 2y = cos 4y
sin 2y = sin (90 - 4y)
2y = 90 - 4y
6y = 90
y =15

Answer 2

Use the trig identity

(sin(2y))^2 = [1-cos(4y)]/2

to eliminate the cos from your eqn

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

You are left with a quadratic eqn for s = sin(2y)

2s^2 + s -1 = 0

This solves to: s = 1/2 or s = -1.
Now go back to y:

sin(2y) = 1/2 gives y = pi/12 = 15 deg
or y = (5/12)pi = 75 deg

sin(2y) = -1 gives y = 3pi/4 = 135 deg

(plus any multiple of 180 degs in each case).

Answer 3

<=> sin2y - cos4y = 0 <=> sin2y - sin(90 - 4y) =0
<=> 2 cos(45-y)*sin(3y - 45) = 0
<=> cos (45-y)=0 or sin(3y-45)=0
* cos(45-y) = 0 then 45 - y = 90 + k * 180 => y = -45 - k*180
so y = 315 , 135
* sin(3y-45) = 0 then 3y-45 = k*180 => y = 15 + k*60
=> y = 15 , 135 , 195 , 255 , 315

then y = 15 , 135 , 195 , 255 , 315

Answer 4

2