И так держи))) график построил и написал решение)
![\cos4x-6\cos 2x\cos x-4\sin^2x +5=0 \\ 2\cos^22x-1-6\cos 2x\cos x-4+4\cos^2x+5=0\\ 2\cos^22x-6\cos2x\cos x+4\cos^2x=0|:2 \\ \cos^22x-3\cos2x\cos x+2\cos^2x=0\\ (2\cos^2x-1)^2-3(2\cos^2x-1)\cos x+2\cos^2x=0](https://tex.z-dn.net/?f=%5Ccos4x-6%5Ccos+2x%5Ccos+x-4%5Csin%5E2x+%2B5%3D0+%5C%5C+2%5Ccos%5E22x-1-6%5Ccos+2x%5Ccos+x-4%2B4%5Ccos%5E2x%2B5%3D0%5C%5C+2%5Ccos%5E22x-6%5Ccos2x%5Ccos+x%2B4%5Ccos%5E2x%3D0%7C%3A2+%5C%5C+%5Ccos%5E22x-3%5Ccos2x%5Ccos+x%2B2%5Ccos%5E2x%3D0%5C%5C+%282%5Ccos%5E2x-1%29%5E2-3%282%5Ccos%5E2x-1%29%5Ccos+x%2B2%5Ccos%5E2x%3D0)
Произведем замену переменных
Пусть
![\cos x=t\,\,\,(|t| \leq 1)](https://tex.z-dn.net/?f=%5Ccos+x%3Dt%5C%2C%5C%2C%5C%2C%28%7Ct%7C+%5Cleq+1%29)
, тогда получаем
![(2t^2-1)^2-3(2t^2-1)t+2t^2=0|:t^2\\((2(t- \frac{1}{2t} ))^2-3(2(t- \frac{1}{2t}))+2=0](https://tex.z-dn.net/?f=%282t%5E2-1%29%5E2-3%282t%5E2-1%29t%2B2t%5E2%3D0%7C%3At%5E2%5C%5C%28%282%28t-+%5Cfrac%7B1%7D%7B2t%7D+%29%29%5E2-3%282%28t-+%5Cfrac%7B1%7D%7B2t%7D%29%29%2B2%3D0)
Пусть
![t- \frac{1}{2t}=z](https://tex.z-dn.net/?f=t-+%5Cfrac%7B1%7D%7B2t%7D%3Dz)
, тогда получаем
![(2z)^2-3\cdot 2z+2=0 \\ 4z^2-6z+2=0|:2 \\2z^2-3z+1=0 \\ D=b^2-4ac=(-3)^2-4\cdot 2\cdot 1=1 \\ z_1= \frac{3-1}{2\cdot 2} =0.5 \\ z_2= \frac{3+1}{2\cdot 2}=1](https://tex.z-dn.net/?f=%282z%29%5E2-3%5Ccdot+2z%2B2%3D0+%5C%5C+4z%5E2-6z%2B2%3D0%7C%3A2+%5C%5C2z%5E2-3z%2B1%3D0+%5C%5C+D%3Db%5E2-4ac%3D%28-3%29%5E2-4%5Ccdot+2%5Ccdot+1%3D1+%5C%5C+z_1%3D+%5Cfrac%7B3-1%7D%7B2%5Ccdot+2%7D+%3D0.5+%5C%5C+z_2%3D+%5Cfrac%7B3%2B1%7D%7B2%5Ccdot+2%7D%3D1+)
Возвращаемся к замене
![t- \frac{1}{2t}=0.5|\cdot 2t \\ 2t^2-t-1=0 \\ D=b^2-4ac= (-1)^2-4\cdot 2\cdot (-1)=9 \\ t_1= \frac{1-3}{2\cdot2}=-0.5 \\ t_2= \frac{1+3}{2\cdot 2}=1](https://tex.z-dn.net/?f=t-+%5Cfrac%7B1%7D%7B2t%7D%3D0.5%7C%5Ccdot+2t+%5C%5C+2t%5E2-t-1%3D0+%5C%5C+D%3Db%5E2-4ac%3D+%28-1%29%5E2-4%5Ccdot+2%5Ccdot+%28-1%29%3D9+%5C%5C+t_1%3D+%5Cfrac%7B1-3%7D%7B2%5Ccdot2%7D%3D-0.5+%5C%5C+t_2%3D+%5Cfrac%7B1%2B3%7D%7B2%5Ccdot+2%7D%3D1++)
![t- \frac{1}{2t}=1|\cdot 2t \\ 2t^2-2t-1=0 \\ D=b^2-4ac=(-2)^2-4\cdot 2\cdot (-1)=12;\,\, \sqrt{D} =2 \sqrt{3} \\ t_1= \frac{2-2 \sqrt{3} }{2\cdot 2} = \frac{1- \sqrt{3} }{2}](https://tex.z-dn.net/?f=t-+%5Cfrac%7B1%7D%7B2t%7D%3D1%7C%5Ccdot+2t+%5C%5C+2t%5E2-2t-1%3D0+%5C%5C+D%3Db%5E2-4ac%3D%28-2%29%5E2-4%5Ccdot+2%5Ccdot+%28-1%29%3D12%3B%5C%2C%5C%2C+%5Csqrt%7BD%7D+%3D2+%5Csqrt%7B3%7D++%5C%5C+t_1%3D+%5Cfrac%7B2-2+%5Csqrt%7B3%7D+%7D%7B2%5Ccdot+2%7D+%3D+%5Cfrac%7B1-+%5Csqrt%7B3%7D+%7D%7B2%7D+)
![t_2= \frac{2+2 \sqrt{3} }{2\cdot 2} = \frac{1+ \sqrt{3} }{2}](https://tex.z-dn.net/?f=t_2%3D+%5Cfrac%7B2%2B2+%5Csqrt%7B3%7D+%7D%7B2%5Ccdot+2%7D+%3D+%5Cfrac%7B1%2B+%5Csqrt%7B3%7D+%7D%7B2%7D+)
- не удовлетворяет условие при |t|≤1.
Обратная замена
![\cos x=-0.5 \\ x=\pm \frac{2 \pi }{3} +2 \pi n,n \in Z \\ \\ \cos x=1 \\ x=2\pi n,n \in Z \\ \\ \cos x= \frac{1- \sqrt{3} }{2} \\ x=\pm \arccos( \frac{1- \sqrt{3} }{2} )+2 \pi n,n \in Z](https://tex.z-dn.net/?f=%5Ccos+x%3D-0.5+%5C%5C+x%3D%5Cpm+%5Cfrac%7B2+%5Cpi+%7D%7B3%7D+%2B2+%5Cpi+n%2Cn+%5Cin+Z+%5C%5C++%5C%5C+%5Ccos+x%3D1+%5C%5C+x%3D2%5Cpi+n%2Cn+%5Cin+Z+%5C%5C+%5C%5C+%5Ccos+x%3D+%5Cfrac%7B1-+%5Csqrt%7B3%7D+%7D%7B2%7D++%5C%5C+x%3D%5Cpm+%5Carccos%28+%5Cfrac%7B1-+%5Csqrt%7B3%7D+%7D%7B2%7D+%29%2B2+%5Cpi+n%2Cn+%5Cin+Z)
Наши действия: надо выразить из данной записи х, который зависит от у ( у нас пока у зависит от х)
А потом буквы х и у поменять местами, т.к. функцию всё-таки буквой у обозначают...
начали.
у = 2·(х + 6)^-1
y = 2/(x + 6)
х + 6 = 2/у
х = 2/у -6
х = (2 - 6у)/у
Теперь конечный результат:
у = (2- 6х)/х
x2-25-x-x+1=9x затем x2-11X-24=0 potom ищешь дискпиминант и корни