Для решения данного уравнения, следует использовать тригонометрические тождества.

cos^2(2x) = (1 + cos(4x))/2

sin^4x + cos^4x = (sin^2x + cos^2x)^2 - 2sin^2xcos^2x = 1 - 2sin^2xcos^2x

Итак, уравнение примет вид:

1 - 2sin^2x*cos^2x = (1 + cos(4x))/2

Учитывая, что sin(2x) = 2sinxcosx, мы можем преобразовать уравнение:

1 - sin^2(2x)/2 = (1 + cos(4x))/2
2 - sin^2(2x) = 1 + cos(4x)
sin^2(2x) + cos(4x) = 1

cos(4x) = 1 - sin^2(2x)

Taking into account that cos(2x) = cos^2(x) - sin^2(x) and sin(2x) = 2sin(x)cos(x), we have:

cos(4x) = cos^2(2x) - sin^2(2x) = (cos^2(x) - sin^2(x))^2 - (2sin(x)cos(x))^2 = cos^4(x) - 2cos^2(x)sin^2(x) + sin^4(x) - 4sin^2(x)cos^2(x) = 1 - 2sin^2(x) + sin^4(x) - 4sin^2(x)(1 - sin^2(x)) = 1 - 2sin^2(x) + sin^4(x) - 4sin^2(x) + 4sin^4(x) = 5sin^4(x) - 6sin^2(x) + 1

Thus, the equation becomes:

5sin^4(x) - 6sin^2(x) + 1 = 1

5sin^4(x) - 6sin^2(x) = 0

sin^2(x)(5sin^2(x) - 6) = 0

sin^2(x) = 0 or 5sin^2(x) = 6

sin(x) = 0 or sin(x) = ±√(6/5)

Therefore, the solutions to the equation sin^4x + cos^4x = cos^2(2x) are:

x = kπ, where k is an integer, or x = arcsin(±√(6/5)) + 2πn, where n is an integer.