4x²+12x+9-4(1-2x+x²)=1
4x²+12x+9-4+8x-4x²=1
20x+5=1
20x=1-5
20x=-4
x=-0.2
А)
![4cos^4x-3cos(2x)-1=0 \\ 4cos^4x-3(cos^2x-sin^2x)-1=0 \\ 4cos^4x-3(cos^2x-1+cos^2x)-1=0 \\ 4cos^4x-6cos^2x+2=0 \\ 2cos^4x-3cos^2x+1=0 \\ D=9-4(2)=1 \\ cos^2x= \frac{3+-1}{4} \\ cos^2x=1; cos^2x=1/2 \\ x_{1}= \pi n; x_{2} =+- \frac{ \pi }{4} + \pi n](https://tex.z-dn.net/?f=4cos%5E4x-3cos%282x%29-1%3D0+%5C%5C+4cos%5E4x-3%28cos%5E2x-sin%5E2x%29-1%3D0+%5C%5C+4cos%5E4x-3%28cos%5E2x-1%2Bcos%5E2x%29-1%3D0+%5C%5C+4cos%5E4x-6cos%5E2x%2B2%3D0+%5C%5C+2cos%5E4x-3cos%5E2x%2B1%3D0+%5C%5C+D%3D9-4%282%29%3D1+%5C%5C+cos%5E2x%3D+%5Cfrac%7B3%2B-1%7D%7B4%7D++%5C%5C+cos%5E2x%3D1%3B+cos%5E2x%3D1%2F2+%5C%5C+x_%7B1%7D%3D+%5Cpi+n%3B++x_%7B2%7D+%3D%2B-+%5Cfrac%7B+%5Cpi+%7D%7B4%7D+%2B+%5Cpi+n)
б)x∈[-7pi/2; -2pi]
Из первой группы корней
![x=-3 \pi](https://tex.z-dn.net/?f=x%3D-3+%5Cpi+)
, из второй группы корней
![x=-2 \pi - \frac{ \pi }{4} =- \frac{9 \pi }{4}](https://tex.z-dn.net/?f=x%3D-2+%5Cpi+-+%5Cfrac%7B+%5Cpi+%7D%7B4%7D+%3D-+%5Cfrac%7B9+%5Cpi+%7D%7B4%7D+)
,
![x= -2 \pi - \frac{3 \pi }{4} = -\frac{ 11\pi }{4}](https://tex.z-dn.net/?f=x%3D+-2+%5Cpi+-+%5Cfrac%7B3+%5Cpi+%7D%7B4%7D+%3D+-%5Cfrac%7B+11%5Cpi+%7D%7B4%7D+)
,
![x=-3 \pi - \frac{ \pi }{4} =- \frac{13 \pi }{4}](https://tex.z-dn.net/?f=x%3D-3+%5Cpi+-+%5Cfrac%7B+%5Cpi+%7D%7B4%7D+%3D-+%5Cfrac%7B13+%5Cpi+%7D%7B4%7D+)
Ответ: а)
![x_{1}= \pi n; x_{2} =+- \frac{ \pi }{4} + \pi n](https://tex.z-dn.net/?f=x_%7B1%7D%3D+%5Cpi+n%3B+x_%7B2%7D+%3D%2B-+%5Cfrac%7B+%5Cpi+%7D%7B4%7D+%2B+%5Cpi+n)
, где n-целое число
б)
![- \frac{13 \pi }{4}; -\frac{ 11\pi }{4}; - \frac{9 \pi }{4}](https://tex.z-dn.net/?f=-+%5Cfrac%7B13+%5Cpi+%7D%7B4%7D%3B+-%5Cfrac%7B+11%5Cpi+%7D%7B4%7D%3B+-+%5Cfrac%7B9+%5Cpi+%7D%7B4%7D)
Под буквой б) нужно построить окружность и поставить нужные точки
Вот так все делается в первом уравнение 24, во втором 1/50
При х = 6
4/18 = 9 - 1/12
4/18 = 8/12
1/9 = 4/3
при х = 8
4/24 = 9 -1/16
4/24 = 8/16
1/3 = 1/4