sin(x/2- π/6)+1=0
1 = sin(π/2)
sin(x/2 - π/6) + sin(π/2) = 0
sin(x/2 - π/6) = -sin(π/2)
sin(x/2 - π/6) = sin(-π/2)
x/2 - π/6 = (3x - π)/6
(3x - π)/6 = -π/2
3x - π + 3π = 0
3x +2π = 0
3x = -2π
x = -2π/3
Ответ: -2π/3
Y=-1/9x³+3x
D(f)∈R
-1/9x(x²-27)=0
x=0 x=-3√3 x=3√3
(0;0),(-3√3;0),(3√3;0) точки пересечения с осями
f`(x)=-1/3x²+3
-1/3x²=-3
x²=9
x=-3 x=3
_ + _
-----------------(-3)-----------(3)-------------------
убыв min возр max убыв
y(-3)=3-9=-6
y(3)=-3+9=6
2х-20=4-2х
2х+2х=-2х+20
4х=18
х=4,5
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