2x<π/3+πn
x<π/6+πn/2
x∈(-π/2+πn;π/6+πn/)
sin(x/2-π/4)+√2/2<0
sin(x/2-π/4)<-√2/2
5π/4+2πn<sin(x/2-π/4)<7π/4+2πn
5π/4-π/4+2πn<sinx/2<7π/4-π/4+2πn
π+2πn<sinx/2<3π/2+2πn
2π+4πn<sinx<3π+4πn
10x - 12 < 0
x < 12/10 = 5/6
Подходят числа 1; -7; 0; 0,5
(sin(x))^3 - sin(x) - cos(x) = 0
sin(x) * ((sin(x))^2 - 1) - cos(x) = 0
sin(x) * ((1 - (cos(x))^2) - 1) - cos(x) = 0
sin(x) * (cos(x))^2 + cos(x) = 0
cos(x) * (sin(x) * cos(x) + 1) = 0
sin(x) * cos(x) = -1
0.5 * sin(2*x) = -1
sin(2*x) = -2
решения нет
остается:
cos(x) = 0
Cosa=+-корень из (1-9/25)=+-4/5,так как а принадлежит 2четверти , то cosa<0; cosa =-4/5; tga= sina/cosa=3/5:(-4/5)=-3/4;ctga=-4/3
<span>у = 1,4 х - 2,6 и у = 3,4 - 1,6 х
1.4x -2.6 = 3.4 -1.6x
1.4x +1.6x = 3.4 +2.6
3x = 6
x = 6/3
x = 2
Ответ С</span>