Sin(60°+a)sin(60°-a) = (1/2)*[cos(60° + α - 60° + α<span>) -
- cos(</span>60° + α + 60° - α)] = (1/2)*[cos2α - cos(120°)] =
<span>= (1/2)*[cos2α + cos(60°)] = (1/2)*[cos2α + 1/2)] = (1/2)*cos2α + 1/4</span>
B1=2*3^1-1=2*3^0=2*1=2
b2=2*3^2-1=2*3^1=2*3=6
b3=2*3^3-1=2*3^2=2*9=18
b4=2*3^4-1=2*3^3=2*27=54
b5=2*3^5-1=2*3^4=2*81=162
S(bn)=2+6+18+54+162=242
или: Sn=(b1*(q^n-1))/(q-1);q=b2:b1=6:2=3
S5=(2*(3^5-1))/(3-1)=2*242/2=242
Ответ:242
Cosx+3sinx/2=-1
3*√(1-cosx)/2=-(cosx+1)
(1-cosx)/2=(cos²x+2cosx+1)/9
9-9cosx=2cos²x+4cosx+2
2cos²x+13cosx-7=0
cosx=a
2a²+13a-7=0
D=225
a1=(-13+15)/4=1/2 cosx=1/2 x=+-π/3+2πk
a2=(-13-15)/4=-7 нету решение
Розвязане 7, 8 завдання )))))