<span><span><span>Применим формулу замены произведения косинусов их суммой
cos α · cos β</span>
=
(cos(α-β) + cos(α+β))/
</span><span>
2</span></span>
cos44*cos16=1/2(cos60+cos28)=1/2(1/2+cos28), cos60=1/2
cos 59*cos31=1/2(cos90+cos28)=1/2<span>cos28, </span> cos90=0
1/2(1/2+cos28)-1/2<span>cos28=1/4+1/2</span>cos28-1/2cos28=1/4
a₁ = - 6,5 a₂ = - 6 Sₙ = - 42,5 n = ?
a₂ = a₁ + d
d = a₂ - a₁ = - 6 - (- 6,5) = - 6 + 6,5 = 0,5
![S_{n}=\frac{2a_{1}+d(n-1) }{2}*n\\\\-42,5=\frac{2*(-6,5)+0,5(n-1)}{2}*n\\\\(-13+0,5n-0,5)*n=-85\\\\0,5n^{2}-13,5n+85=0\\\\n^{2}-27n+170=0\\\\D=(-27)^{2}-4*170=729-680=49=7^{2}\\\\n_{1}=\frac{27+7}{2}=17\\\\n_{2}=\frac{27-7}{2}=10](https://tex.z-dn.net/?f=S_%7Bn%7D%3D%5Cfrac%7B2a_%7B1%7D%2Bd%28n-1%29%20%7D%7B2%7D%2An%5C%5C%5C%5C-42%2C5%3D%5Cfrac%7B2%2A%28-6%2C5%29%2B0%2C5%28n-1%29%7D%7B2%7D%2An%5C%5C%5C%5C%28-13%2B0%2C5n-0%2C5%29%2An%3D-85%5C%5C%5C%5C0%2C5n%5E%7B2%7D-13%2C5n%2B85%3D0%5C%5C%5C%5Cn%5E%7B2%7D-27n%2B170%3D0%5C%5C%5C%5CD%3D%28-27%29%5E%7B2%7D-4%2A170%3D729-680%3D49%3D7%5E%7B2%7D%5C%5C%5C%5Cn_%7B1%7D%3D%5Cfrac%7B27%2B7%7D%7B2%7D%3D17%5C%5C%5C%5Cn_%7B2%7D%3D%5Cfrac%7B27-7%7D%7B2%7D%3D10)
Ответ : 10 членов
(2х-1)*(3х+4)-6х^2=1
6х^2+5х-4-6х^2=1
5х=5
х=1
(х-3)^2 - корень из 5(x-3)<0
(x-3)( x-3- корень из 5)<0.
x= 3
x= 3+ корень из 5
и решением будет являть внутренняя часть т.к. x<0.
( 3; 3+ корень из 5)