cos^2 x - 2sinxcosx - sin^2 x = 0 [ 0 ≤2x≤ pi ]cos^2 x - 2sinxcosx - sin^2 x = 0 [ 0 ≤2x≤ pi ]
来源：学生作业帮助网 编辑：六六作业网 时间：2024/05/24 01:41:31
cos^2x-2sinxcosx-sin^2x=0[0≤2x≤pi]cos^2x-2sinxcosx-sin^2x=0[0≤2x≤pi]cos^2x-2sinxcosx-sin^2x=0[0≤2x≤p
cos^2 x - 2sinxcosx - sin^2 x = 0 [ 0 ≤2x≤ pi ]cos^2 x - 2sinxcosx - sin^2 x = 0 [ 0 ≤2x≤ pi ]
cos^2 x - 2sinxcosx - sin^2 x = 0 [ 0 ≤2x≤ pi ]
cos^2 x - 2sinxcosx - sin^2 x = 0 [ 0 ≤2x≤ pi ]
cos^2 x - 2sinxcosx - sin^2 x = 0 [ 0 ≤2x≤ pi ]cos^2 x - 2sinxcosx - sin^2 x = 0 [ 0 ≤2x≤ pi ]
原式＝cos2x-sin2x=2^(1/2)*cos(2x+pi/4)=0,因为0
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