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

问题描述:

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