Solve the equation for the interval 0 2 pi sin2 2x 1 0 2 pi

Solve the equation for the interval [0, 2 pi) sin^2 2x = 1 0, 2 pi/3, pi, 4 pi/3 pi/8, 9 pi/8 pi/4/3 pi/4, 5 pi/4, 7 pi/4 no solution

0<=x<=2

0<=2x<=4

sin²2x=1

=> sin2x =1, sin2x =-1

=>2x=_/2,5_/2, 3_/2,7_/2

=>x=_/4,5_/4, 3_/4,7_/4

=>x=_/4,3_/4,5_/4, 7_/4