find the exact solutions of the equation in the interval [0, 2π). (Enter your answers as a comma-separated list.)?

2 Answers

• 4 sin 4x = -8 sin 2x

  4*2sin 2x cos 2x = - 8sin 2x

  or sin 2x ( cos 2x + 1) = 0

  so sin 2x =0 or cos 2x = -1

  or 2x = npi and 2x = npi

  so x = n*pi/2 for n = 0, 1, 2,,3

• 8sin(2x)cos(2x) = -8sin(2x)

  sin(2x)(cos(2x) + 1) = 0

  sin(2x) = 0 ==> 2x = 0, pi, 2pi, 3pi

  cos(2x) = -1 ==> 2x = pi, 3pi

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