Решите уравнение
7/4cos(x/4)=cos³(x/4)+sin(x/4)
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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