Cos(2x+π/9)=-√3/2
2x+π/9=-5π/6+2πn U 2x+π/9=5π/6+2πn
2x=-17π/18+2πn U 2x=13π/18+2πn
x=-17π/36+πn U x=13π/36+πn
y=0 при x=4\5 .y=-12 при x=-1,6. y=1 при x=1 . Если тебе нужен решение скажи
(3x+4y)в квадрате - (4y-3x)в квадрате =
=((4y+3х)-(4y-3x))((4y+3х)+(4y-3x))=
=(4y+3х-4y+3x)(4y+3х+4y-3x)=6x*8y=48ху
2а-8ау²=2а(1-4у²)=2а(1-2у)(1+2у)
Решение
1) cost = √2/2
t = (+ -)arccos(√2/2) + 2πk, k∈z
t = (+ -)(π/4) + 2πk, k∈Z
2) cos(3π + t) = - 1/2
- cost = - 1/2
cost = 1/2
t = (+ -)arccos(1/2) + 2πn, n∈z
t = (+ -)(π/3) + 2πn, n∈Z