3sin(5П/2-а)=3(sin5П/2cosa-sinacos5П/2)=3(sin(2π+π\2)cosa-sinacos(2π+π\2))=
3(sinπ\2cosa-sinacosπ\2).
sin²a+cos²a=1
|cosa|=√1-sin²a=√1-0,64=0,6
cosa=-0,6 так як cosa<0 на проміжку a∈(π;3π\2)
3(sinπ\2cosa-sinacosπ\2)=3*cosa=-3*o,6=-1,8
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Cos6x + 6cos²3x = 1
cos²3x - sin²3x + 6cos²3x = 1
7cos²3x - sin²3x = 1
7cos²3x - (1 - cos²3x) = 1
7cos²3x - 1 + cos²3x = 1
8cos²3x = 2
4cos²3x = 1
cos²3x = 1/4
cos3x = 1/2
cos3x = -1/2
3x = -π/3 + 2πn
3x = π/3 + 2πn
3x = 2π/3 + 2πn
3x = -2π/3 + 2πn
x = -π/9 + 2πn/3
x = π/9 + 2πn/3
x = 2π/9 + 2πn/3
x = -2π/9 + 2πn/3
OTBET: -π/9 + 2πn/3; π/9 + 2πn/3; -2π/9 + 2πn/3; 2π/9 + 2πn/3; n ∈ Z
A1=4,a2=6,a3=8
d=2
a10=a1+9.2=4+18=22
s10=10(4+22)/2=260/2=130
Otvet: 130