Задача 64019 ...
Условие
(√2sin(x/2)-√2cos(x/2))^2=5+sin(pi/4-x/2)
Решение
(√2sin(x/2)–√2cos(x/2))^2=5+(sqrt(2)/2)*cos(x/2)-(sqrt(2)/2)* sin(x/2)
(√2sin(x/2)–√2cos(x/2))^2=5+(sqrt(2)/2)*(cos(x/2)-sin(x/2))
(√2(sin(x/2)–cos(x/2)))^2=5+(sqrt(2)/2)*(cos(x/2)-sin(x/2))
2*[blue](sin(x/2)–cos(x/2))[/blue]^2-(sqrt(2)/2)*[blue](cos(x/2)-sin(x/2))[/blue]-5=0
Замена переменной
[blue]sin(x/2)–cos(x/2)[/blue]=t
2t^2-(sqrt(2)/2)*t-5=0
4t^2-(sqrt(2))*t-10=0
D=2+160=162=(9sqrt(2))^2
t_(1)=(sqrt(2)-9sqrt(2))/8; t_(2)=(sqrt(2)+9sqrt(2))/8
t_(1)=-sqrt(2); t_(2)=5sqrt(2)/4
Обратный переход
[blue]sin(x/2)–cos(x/2)[/blue]=-sqrt(2) ИЛИ [blue]sin(x/2)–cos(x/2)[/blue]=5sqrt(2)/4
Решаем методом введения вспомогательного угла
[blue](1/sqrt(2))*sin(x/2)–(1/sqrt(2))*cos(x/2)[/blue]=-1
sin((x/2)-(π/4))=-1
(x/2)-(π/4)=-(π/2)+2πk, k ∈ Z
(x/2)=-(π/2)+(π/4)+2πk, k ∈ Z
(x/2)=-(π/4)+2πk, k ∈ Z
[b]x=-(π/2)+4πk, k ∈ Z[/b]
[blue]sin(x/2)–cos(x/2)[/blue]=5sqrt(2)/4 - уравнение не имеет корней.
[blue]|sin(x/2)–cos(x/2)|[/blue] ≤ sqrt(2)
5sqrt(2)/4 ≥ sqrt(2)