Đk sin2x 0

Ta có sin22x + sin2x.cos2x = 1 cos2x

1 cos22x + sin2x.cos2x 1 + cos2x = 0 cos2x( cos2x + sin2x + 1) = 0

cos2x = 0 cos2x sin2x = 1

  • cos2x = 0 x = , kZ
  • cos2x sin2x = 1 x = k (loại) x = /4 +  k

2/ cos23x.cos2x cos2x = 0

cos6x.cos2x 1 = 0 (cos8x + cos4x) 1 = 0 cos8x + cos4x = 2

1 2sin24x + 1 2sin22x = 2 sin24x + sin22x = 0

sin2x = 0 x = k/2, k Z

nguon VI OLET