2сos(5x/2)cos(x/2)-2sin(5x/2)cos(5x/2)=0
2cos(5x/2)*(cos(x/2)-sin(5x/2))=0
cos(5x/2)=0⇒5x/2=π/2+πn⇒x=π/5+2πn/5,n∈z
cos(x/2)-sin(5x/2)=0
cos(x/2)-cos(π/2-5x/2)=0
-2sin(3x/2-π/4)sin(-x+π/4)=0
2sin(3x/2-π/4)sin(x-π/4)=0
sin(3x/2-π/4)=0⇒3x/2-π/4=πk⇒3x/2=π/4+πk⇒x=π/6+2πk/3,k∈z
sin(x-π/4)=0⇒x-π/4=πm⇒x=π/4+πm,m∈z
4<a<5
2*4<2a<2*5
8<2a<10
8-7<2a-7<10-7
1<2a-7<3
Ответ: 1<2a-7<3
12^(2k-2)/22^(2k-2)6^(2k-3)=2^(2k-2)6^(2k-2)*6/11^(2k-2)2^(2k-2)6^(2k-2)=6/11^(2k-2)
Обозначим: arccos(3/8) = x ---> cos(x) = 3/8
нужно найти: cos(2x) = 2*cos²(x) - 1 = 2*9/64 - 1 = 9/32 - 32/32 = -23/32
X^2+5x+9=0
D=25-36=-11
D<0
Корней нет