Solve sin² (x) = sin (x)

solve for x between 0 < X < 2π

Show work please

sin x ( sin x - 1 ) = 0

sin x = 0 , sin x = 1

x = 0 , π , 2π , π/2 for 0 ≤ x ≤ 2π

Thank you for rating----pleased to help.

Report Abuse

sin^2(x) = sin(x)

=> sin^2(x) - sin(x) = 0

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

case 1: sin(x) = 0

=> x = Π

case 2: sin(x) - 1 = 0

=> sin(x) = 1

=> x = Π/2

x = Π/2 and Π