Решить уравнение:

2√2 cosx+2-cos2x=0

2√2cosx + 2 - (2cos²x - 1) = 0

2cos²x - 2√2cosx - 3 = 0

cosx = t

2t² - 2√2t - 3 = 0

D/4 = (b/2) ² - ac = 2 + 6 = 8

t = (-b/2 + - √ (D/4)) / a

t = (√2 - 2√2) / 2 = - √2/2 t = (√2 + 2√2) / 2 = 3√2/2

cosx = - √2/2 cosx = 3√2 / 2 > 1 нет корней

x = + - arccos ( - √2/2) + 2πn

x = + - (π - arccos (√2/2)) + 2πn

x = + - 3π/4 + 2πn, n∈Z