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