7/4cos (x/4) = cos^3 (x/4) + sin (x/2)

По формуле синуса двойного угла

7/4*cos (x/4) = cos^3 (x/4) + 2sin (x/4) * cos (x/4)

cos^3 (x/4) + cos (x/4) * (2sin (x/4) - 7/4) = 0

cos (x/4) * (cos^2 (x/4) + 2sin (x/4) - 7/4) = 0

1) cos (x/4) = 0; x/4 = pi/2 + pi*k; x1 = 2pi + 4pi*k

2) 1 - sin^2 (x/4) + 2sin (x/4) - 7/4 = 0

Умножаем все на - 1 и делаем замену sin (x/4) = y

y^2 - 2y + 7/4 - 1 = 0

y^2 - 2y + 3/4 = 0

D/4 = 1 - 3/4 = 1/4 = (1/2) ^2

y1 = sin (x/4) = 1 - 1/2 = 1/2; x/4 = (-1) ^n*pi/6 + pi*n; x2 = (-1) ^n*2pi/3 + 4pi*n

y2 = sin (x/4) = 1 + 1/2 = 3/2 - решений нет, потому что sin x < = 1

Ответ: x1 = 2pi + 4pi*k; x2 = (-1) ^n*2pi/3 + 4pi*n