Решите уравнение : cos^2x-cosx=sin^2x

Sin^2x = 1 - cos^2x, подставляем, cos^2(x) - cos(x) = 1 - cos^2(x); 2*cos^2(x) - cos(x) - 1 = 0; cos(x) = t; (-1)<=t<=1, 2*t^2 - t - 1= 0; D = 1 + 4*2 = 9 = 3^2; t1 = (1-3)/4 = -2/4 = -1/2; t2 = (1+3)/4 = 1, cos(x) = -1/2; x = (п-(п/3)) + 2*п*m, = (2/3)*п + 2*п*m, или x = -(2/3)*п + 2пn, cos(x) = 1, <=> x = 2пk.