cosx 1 2 cos 2x 1

Ответ от Sheogorath[гуру]
cos(x) + 2(cos^2(x) - sin^2(x)) = 1 cos(x) + 2cos^2(x) - 2sin^2(x) = cos^2(x) + sin^2(x) cos(x) + cos^2(x) - 3sin^2(x) = 0 cos(x) + cos^2(x) - 3(1-cos^2(x)) = 0 cos(x) + cos^2(x) - 3 + 3cos^2(x) = 0 4cos^2(x) + cos(x) - 3 = 0 ]cos(x) = t 4t^2 + t - 3 = 0 D = 1 + 48 = 49 t = (-1+-7)/8 = -1 ; 3/4. cos(x) = Pi + 2*Pi*k, k - целое cos(x) = +- arccos(3/4) + 2*Pi*n, n - целое.