Ответы на вопрос
Ответил Матов
0
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%0A8cos%5E3x-6cosx-1%3D0%5C%0Acosx%288cos%5E2x-6cosx%29-1%3D0%5C%0A2cos3x-1%3D0%5C%0A2cos3x%3D1%5C%0Ax%3Dfrac%7Bpi%7D%7B9%7D%5C%0Ax%3Dcos%28frac%7Bpi%7D%7B9%7D%7D%29%3D0.939%5C%0Avtoroy++koren%27+posle++delenia++nawego++vyrazhenia++na++cos%28frac%7Bpi%7D%7B9%7D%7D%29%5C%0A%0A "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}})\")
\
teper'\ zamena \ na \ x=sinx\
![teper'\ zamena \ na \ x=sinx\](https://tex.z-dn.net/?f=teper%27%5C+zamena+%5C+na+%5C+x%3Dsinx%5C%0A "teper'\ zamena \ na \ x=sinx\")
8sin^3x-6sinx-1=0
-2sin3x-1=0
x=-pi/18+pi*k/2
sin(-pi/18)=-0.17...
sin(-5pi/18)=-0.766....
teper'\ zamena \ na \ x=sinx\
8sin^3x-6sinx-1=0
-2sin3x-1=0
x=-pi/18+pi*k/2
sin(-pi/18)=-0.17...
sin(-5pi/18)=-0.766....
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