cos^2 x - 2sinxcosx - sin^2 x = 0 [ 0 ≤2x≤ pi ]cos^2 x - 2sinxcos_数学解答

cos^2 x - 2sinxcosx - sin^2 x = 0 [ 0 ≤2x≤ pi ]
cos^2 x - 2sinxcosx - sin^2 x = 0 [ 0 ≤2x≤ pi ]

人气：498 ℃　时间：2020-06-16 17:43:45

解答

原式＝cos2x-sin2x=2^(1/2)*cos(2x+pi/4)=0,因为0