Find all of the exact solutions to the equation in the inter

Find all of the exact solutions to the equation in the interval [0,2 pi). 1 - sin(2x) = cos(4x)

0<=x<=2pi

0<=2x<=4pi

1-sin(2x)=cos(4x)

1-sin(2x)=cos(2*2x)

cos2A=1-2sin²A

1-sin(2x)=1-2sin²(2x)

2sin²(2x)-sin(2x)=0

sin(2x)[2sin(2x) -1]=0

sin2x=0 ,sin2x =1/2

sin2x=0 => 2x =0 ,pi,2pi,3pi

=>x=0,pi/2,pi,3pi/2

sin2x =1/2 =>2x=pi/6 ,5pi/6 ,13pi/6,17pi/6

x=pi/12,5pi/12,13pi/12,17pi/12

therefore x =0,pi/2,pi,3pi/2,pi/12,5pi/12,13pi/12,17pi/12 are solutions