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