Sin(3x+π/4)=1
(sinπ/2=1)
3x+π/4=π/2+2k.π
3x=π/4+2k.π
x=π/12+2k.π/3 , k∈Z
k=0: x=π/12
k=1:x=π/12+2π/3=9π/12=3π/4
k=2: x=π/12+4π/3=17π/12
k=3: x=π/12+6π/3=π/12+2π - yže net v intervale /0,2π
Otvet: x=π/12, x=3π/4, x=17π/12
в 20 ответ равен 1280,19 не знаю
2(x+y)=46
xy = 14400
2x+2y=46
2x=46-2y
x=23-y
(23-y)y=14400
23y-y2-14400=0, а дальше через дискриминант
2А³+А²-17-3А²+А³-А-80=2А³+А³+А²-3А²-А-17-80=3А³-2А²-А-97