Sin^6 x + cos^6 x = cos 2x

Будут использованы следующие формулы:

(a + b) 2 = a^2 + 2ab + b^2; a^3 + b^3 = (a + b) (a^2 - ab + b^2); sin^2α + cos^2α = 1; 1 - 2sin^2α = cos2α; sin^6x + cos^6x = cos2x; (sin^2x) ^3 + (cos^2x) ^3 = cos2x; (sin^2x + cos^2x) (sin^4x - sin^2x * cos^2x + cos^4x) = cos2x; (sin^2x + cos^2x) ^2 - 3sin^2x = cos2x; 1 - 3sin^2x = cos2x; 2 - 6sin^2x = 2cos2x; 3 - 6sin^2x - 1 = 2cos2x; 3 (1 - 2sin^2x) - 1 = 2cos2x; 3cos2x - 1 = 2cos2x; 3cos2x - 2cos2x = 1; cos2x = 1; 2x = 2πk, k ∈ Z; x = πk, k ∈ Z.

Ответ: πk, k ∈ Z.