Найти область определения функции: y=arcsin(lg((x-1)/10))

y=arcsin(lg((x-1)/10))

найдем область определения:

-1< = lg(x-1)/10 < =1

0.1< =(x-1)/10< =10

1< = x-1 < =100

2 < =x < = 101

ответ: x e [2; 101]

 

 

 

 

 

 

 

 

 

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

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

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

Объяснение: