Решение
<span>sin²x/4-cos²x/4=1/2
- (cos</span>²x/4 - sin²x/4) = 1/2
cos<span>²x/4 - sin²x/4 = - 1/2
</span>cos[2*(x/4)] = - 1/2
cosx/2 = - 1/2
x/2 = +-arccos(-1/2) + 2πk, k ∈ Z
x/2 = +- [π - arccos(1/2)] + 2πk, k ∈ Z
x/2 = +- [π - π/3<span>)] + 2πk, k ∈ Z
</span>x/2 = +- [2π/3<span>)] + 2πk, k ∈ Z
</span>x = +- [4π/3<span>)] + 4πk, k ∈ Z</span>
3(х+2)=4+2х
3х+6=4+2х
3х-2х=4-6
х=-2
bn = b1*q^(n-1)
b5 = b1*q^4 = 4*(-3)^4= 4*81=324 - ОТВЕТ (5-ый член геом. прогрессии)