Question
Solve for x on the interval (0, 2pi]
(sin2x + cos2x)^2 = 1
(sin2x + cos2x)^2 = 1
Answers
Reiny
sin^2 2x + 2sin(2x) cos(2x) + cos^2 2x = 1
1 + sin (4x) = 1
sin 4x = 0
4x = 0, π, 2π, 3π ....
x = 0 , π/4, π/2, 3π/4, π , 5π/4, 3π/2, 7π/4, 2π
choose the answers according to your notation (0,2π]
1 + sin (4x) = 1
sin 4x = 0
4x = 0, π, 2π, 3π ....
x = 0 , π/4, π/2, 3π/4, π , 5π/4, 3π/2, 7π/4, 2π
choose the answers according to your notation (0,2π]