6
x + y = П/4
sinx/cosx + siny/cosy = 1 | x,y <> П/2 + Пk
sinx*cosy + siny*cosx = cosx*cosy
sin(x+y) = cosx*cosy
cosx*cosy = sin(П/4)
cosx*cos(П/4-x) = sin(П/4)
cosx*(cos(П/4)*cos(x) + sin(П/4)*sin(x)) = sin(П/4) | cos(П/4) = sin(П/4)
cosx*(cosx+sinx) = 1
cos^2x + cosx*sinx = 1
cosx*sinx - sin^2x = 0
sinx*(cosx - sinx) = 0
sinx = 0 -> x = Пk, y = П/4 - Пk
cosx = sinx -> x = П/4 - Пk, y = Пk
7
cos^2x = sinx*siny
sin^2x = cosx*cosy
1 = sinx*siny + cosx*cosy
1 = cos(x-y)
x-y = П/2 + 2Пk, y = x + П/2 + 2Пk
cos^2x = sinx*sin(x+П/2) = sinx*cosx -> cosx = 0 | cosx = sinx
sin^2x = cosx*cos(x+П/2) = cosx*(-sinx) -> sinx = 0 | sinx = -cosx
--> cosx = 0 | sinx = 0 --> x = Пn/2, y = П(n+1)/2 + 2Пk
8
cosx*sqrt(cos2x) = 0 | cos2x >= 0
2sin^2x = cos(2y-П/3) | 2sin^2x <= 1
cosx*sqrt(cos^2x - sin^2x) = 0
cosx*sqrt(1 - 2sin^2x) = 0
cosx*sqrt(1 - cos(2y-П/3)) = 0
cosx = 0 -> x = П/2 + Пk - > 2sin^2x > 1 - не подходит
cos(2y-П/3) = 1 - > 2y - П/3 = П/2 + 2Пk -> y = 5П/12 + Пk | cos2x = 1 - 2sin^2x = 1 - cos(2y-П/3) = 0 -> x = П/4 + Пn/2
--> x = П/4 + Пn/2, y = 5П/12 + Пk/2
Ответ...........................
напиши готовое домашнее задание и геометрия и там кто создатель этого учебника в инет (глз
Q=√(b4/b2)=√(24/6)=√4=2
b1=(b2)/q=6/2=3
b10=(b1)·q⁹=3·2⁹=3·512=1536
