Задача 62095 ...

6035fc7016e9eb62f3f77539

11.12.2021 17:30:03

Решите неравенство :
1) sinx+cosx>-√2
2) sin(x-1)≤-√3/2

5f3ea7e3faf909182968ddd9

12.12.2021 12:25:36

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1) sinx+cosx>–√2

Делим на √2

(1/√2)* sinx+(1/√2)*cosx>–1

Вспомогательный угол (π/4)

(cos(π/4))* sinx+(sin(π/4))*cosx>–1

sin(x+(π/4))>-1

верно для всех х, кроме sin(x+(π/4))=-1 ⇒ x+(π/4)=-(π /2)+2πn, n ∈[b] Z[/b]

x=-(π/4)-(π /2)+2πn, n ∈[b] Z[/b]

x=-(3π/4)+2πn, n ∈[b] Z[/b]

О т в е т. x ≠ -(3π/4)+2πn, n ∈[b] Z[/b]


2) sin(x–1)≤–√3/2

-(2π/3)+2πn ≤ x-1 ≤ -(π/3)+2πn , n ∈ [b]Z[/b] ( см. рис.)

1-(2π/3)+2πn ≤ x ≤1 -(π/3)+2πn , n ∈ [b]Z[/b]

О т в е т. 1-(2π/3)+2πn ≤ x ≤1 -(π/3)+2πn , n ∈ [b]Z[/b]