Решите неравенство методом интервалов:

1)

1). 1.=5x^2+6x-3-2x^2+3x+4=3x^2+9x+1
     2.=3x^4-12x^2+18x
    3.=28a^2b+24ab^2+2a^2b-16ab^2=30a^2+8ab^2
2).=12m+20m^2-60m-20m^2=-48m
m=-0.2
-48*(-0.2)=9.6
3).1.=5a(a-4b)
    2.=7x^3(1-2x^2)
    3.=2ab(3ab-4a+6b)
4).1.x^2-3x=0
x(x-3)=0
x=0         или     x-3=0
                          x=3
    2.(x-2)(x+5)=0
x-2=0         или       x+5=0
x=2                         x=-5
3).(18xy+6x)+(-24y-8)=6x(3y+1)-8(3y+1)=(3y+1)(6x-8)
(3*0,45+1)(6*5/3-8)=2,35*2=4,7
4).1.=3(a-b)+x(a-b)=(a-b)(3+x)
    2.=(a+b)^2+(3a+3b)=(a+b)^2+3(a+b)=(a+b)(a+b+3)
    3.=(x^8-4X^5)+(X^3-4)=X^5(X^3-4)+(X^3-4)=(x^3-4)(x^5+1)

Sin²2x=3/4
a)  sin2x = -√3/2
2x = (-1)^n*arcsin(-√3/2) + πn, n∈Z
2x = (-1)^(n+1)*arcsin(√3/2) + πn, n∈Z
2x = (-1)^(n+1)*(π/3) + πn, n∈Z
x1 = (-1)^(n+1)*(π/6) + (πn)/2, n∈Z
b)  sin2x = √3/2
2x = (-1)^(n)*arcsin(√3/2) + πk, n∈Z
2x = (-1)^(n)*(π/3) + πk, k∈Z
x2 = (-1)(n)*(π/6)  + (πk)/2, k∈Z

2) 3cosX + 5sin X/2 + 1 = 0
3*(1 - 2sin^2(x/2) + 5sin(x/2) + 1 = 0
 6sin^(x/2) - 5sin(x/2) - 4 = 0
D = 25 + 4*6*4 = 121
a) sin(x/2) = (5 - 11)/12
sin(x/2) = (-1/2)
x/2 =(-1)^(n)* arcsin(-1/2) + πn, n∈z
x/2 = (-1)^(n+1)*(π/6) + πn, n∈Z
x1 = (-1)^(n+1)*(π/3) + πn, n∈z
b)  sin(x/2) = (5 + 11)/12
sin(x/2) = 1
x/2 = π/2 + 2πk, k∈Z
x2 = π + 4πk, k∈z