Задача 46042 ...

marinichvika15

2020-04-05 19:22:30

sin4x/5 cos4x/5 ≥ - 1/4

sova

2020-04-05 19:33:20

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Формула:

sin2 α =2sin α *cos α ⇒ sin α *cos α =(1/2)sin2 α


sin4x/5 cos4x/5=(1/2)sin(8x/5)


(1/2)sin(8x/5) ≥ -1/4


sin(8x/5) ≥ -1/2


-(π/6)+2πn < 8x/5 < (-5π/6)+2π+2πn, n ∈ Z

-(π/6)+2πn < 8x/5 < (7π/6)+2πn, n ∈ Z

Делим на (8/5) или умножаем на (5/8)



- (5/8)*(π/6)+2*(5/8)πn < x < (5/8)*(7π/6)+2*(5/8)πn, n ∈ Z

[b]-(5π/48)+(5/4)*π*n < x < (35π/48)+(5/4)*π*n, n ∈ Z[/b]