Решите неравенство : (2x+1,5)/9> 1/5*(x-1)+x/5

Cos(2x-2π/3) +5sin(x-π/3)+2=0cos2(x-π/3) +5sin(x-π/3)+2=0cos²(x-π/3) - sin²(x-π/3) +5sin(x-π/3)+2=01-sin²(x-π/3) -sin²(x-π/3) +5sin(x-π/3) +2 =01 -2sin²(x-π/3) +5sin(x-π/3) +2=02sin²(x-π/3) - 5sin(x-π/3) -3 =0sin(x-π/3)=y2y² - 5y -3=0D=25 +24=49y₁=5 -7 = -1/2       4y₂ =5+7 =3        4
При у= -1/2sin(x -π/3) = -1/2x-π/3 =(-1)^(n+1) * (π/6) +πn, n∈Zx=(-1)^(n+1) * (π/6) + π/3 +πn, n∈Z
При у=3sin(x-π/3)=3Так как 3∉[-1; 1], то уравнение не имеет корней.
Ответ: (-1)^(n+1) * (π/6) +π/3 +πn, n∈Z.

В) (x + 12)(3 - x) > 0
(x + 12)( x - 3) < 0
    +             -                +
_____₀________₀__________
         - 12            3
x ∈ (- 12 ; 3)
г) (6 + x)(3x - 1) ≤ 0
3(x + 6)(x - 1/3) ≤ 0
(x + 6)(x - 1/3) ≤ 0
       +                 -                  +
___________________________
               - 6                 1/3
x ∈ [- 6 ; 1/3]

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