Найти корни уравнения cos(x+(π/4))=0, принадлежащие отрезку

[−2π;−π/2]

Объяснение:

1) cos(x+(π/4))=0

x+(π/4)=π/2+2πn  или  x+(π/4)= -π/2+2πm

x=-π/4+π/2+2πn  или  x=-π/4-π/2+2πm

x=π/4+2πn            или  x=-3π/4+2πm где  n∈Z, m∈Z

2)а) -2π ≤ π/4+2πn  ≤-π/2

-2π-π/4 ≤ π/4-π/4+2πn  ≤-π/2-π/4

-9π/4≤2πn≤-3π/4 |:2π

-9/8≤n≤-3/8 ⇒ n=-`1 , х=π/4-2π=-7π/4

б) -2π ≤ -3π/4+2πm  ≤-π/2

-2π+3π/4 ≤ 3π/4-3π/4+2πm  ≤-π/2+3π/4

-5π/4≤2πm≤π/4 |:2π

-5/8≤m≤1/8 ⇒ m=0, х=-3π/4-2π*0=-3π/4.