Решите уравнение 4sin^4 2x+3cos4x-1=0, на отрезке {п; 3п/2}

4sin^4(2x) + 3cos(4x) - 1 = 0

2sin^4(2x) - 3sin^2(2x) + 1 = 0 --> sin^2(2x) = 1; sin^2(2x) = 1/2 --> sin(2x) = ±1; sin(2x) = ±1/√2

--> x = π/4 + πk/2; x = π/8 + πk; x = 3π/8 + πk; x = -π/8 + πk; x = -3π/8 + πk --> x = π/4 + πk/2; x = π/8 + πk/4, k - целое.

Корни, принадлежащие отрезку:

x = 9π/8; x = 5π/4; x = 11π/8.