решите уравнение f ' (x) = 0

f (x) = √2 * (1-x) + cos 2x

F' (x) = 0

f' (x) = (√2 * (1-x) + cos 2x) ' = √2 * (-1) - sin 2x * (2x) '=-√2 - 2sin 2x

-√2 - 2 sin 2x = 0

sin 2x = - √2/2

1) 2x = - π/4 + 2πn

x₁ = - π/8 + πn

2) 2x = π - (-π/4) + 2πk

2x = 5π/4 + 2πk

x₂ = 5π/8 + πk

Ответ: x₁ = - π/8 + πn; x₂ = 5π/8 + πk; n, k∈Z