Решение приведено во вложениях.
-45 = - п/4
-68 = - 25п/66
Сделаем замену
![x=cosx\\ 8cos^3x-6cosx-1=0\\ cosx(8cos^2x-6cosx)-1=0\\ 2cos3x-1=0\\ 2cos3x=1\\ x=\frac{\pi}{9}\\ x=cos(\frac{\pi}{9}})=0.939\\ vtoroy\ koren' posle \ delenia \ nawego \ vyrazhenia \ na \ cos(\frac{\pi}{9}})\\ ](https://tex.z-dn.net/?f=x%3Dcosx%5C%5C%0A8cos%5E3x-6cosx-1%3D0%5C%5C%0Acosx%288cos%5E2x-6cosx%29-1%3D0%5C%5C%0A2cos3x-1%3D0%5C%5C%0A2cos3x%3D1%5C%5C%0Ax%3D%5Cfrac%7B%5Cpi%7D%7B9%7D%5C%5C%0Ax%3Dcos%28%5Cfrac%7B%5Cpi%7D%7B9%7D%7D%29%3D0.939%5C%5C%0Avtoroy%5C%20%20koren%27%20posle%20%5C%20delenia%20%5C%20nawego%20%5C%20vyrazhenia%20%5C%20na%20%5C%20cos%28%5Cfrac%7B%5Cpi%7D%7B9%7D%7D%29%5C%5C%0A%0A)
\\
teper'\\ zamena \\ na \\ x=sinx\\
![teper'\\ zamena \\ na \\ x=sinx\\ ](https://tex.z-dn.net/?f=teper%27%5C%5C%20zamena%20%5C%5C%20na%20%5C%5C%20x%3Dsinx%5C%5C%0A)
8sin^3x-6sinx-1=0
-2sin3x-1=0
x=-pi/18+pi*k/2
sin(-pi/18)=-0.17...
sin(-5pi/18)=-0.766....
Четверть это типо 25%
составляем пропорцию
25%-?
75%-18
(18*25):75=6
5m-0,5m^2-0,5m^2+0,05m^3=0,5m(10-m)-0,05m^2(10-m)=(0,5m-0,05m^2)(10-m)=0,05m(10-m)(10-m)=0,05m(10-m)^2
((12√5)²)/30= (144*5)/30=720/30=24.