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
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cos2x = 0 x = , kZ
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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