Відповідь:1 і -3
Пояснення: y=x^2+2*x-3;
x^2+2*x-3=0;
(a=1, b=2, c=(-3));
D=b^2-4*a*c=2^2-4*1*(-3)=4+12=16;
D>0
x1= -2 + 4/2*1= 2/2 =1
x2= -2 - 4/2*1= -6/2= -3
Нулі функціï: 1 і -3.
б)
![cos(2arctg1)](https://tex.z-dn.net/?f=cos%282arctg1%29)
arctg1 - это значение угла, при котором тангенс равен 1
![arctg1 = \frac{\pi}{4}](https://tex.z-dn.net/?f=arctg1+%3D+%5Cfrac%7B%5Cpi%7D%7B4%7D)
Проверим это
![tg\frac{\pi}{4} = \frac{sin\frac{\pi}{4}}{cos\frac{\pi}{4}} = \frac{\sqrt2}{2} \div \frac{\sqrt2}{2} = \frac{\sqrt2}{2} * \frac{2}{\sqrt2} = 1](https://tex.z-dn.net/?f=tg%5Cfrac%7B%5Cpi%7D%7B4%7D+%3D+%5Cfrac%7Bsin%5Cfrac%7B%5Cpi%7D%7B4%7D%7D%7Bcos%5Cfrac%7B%5Cpi%7D%7B4%7D%7D+%3D+%5Cfrac%7B%5Csqrt2%7D%7B2%7D+%5Cdiv+%5Cfrac%7B%5Csqrt2%7D%7B2%7D+%3D+%5Cfrac%7B%5Csqrt2%7D%7B2%7D+%2A+%5Cfrac%7B2%7D%7B%5Csqrt2%7D+%3D+1)
![cos(2arctg1) = cos(2\cdot \frac{\pi}{4}) = cos\frac{\pi}{2} = 0](https://tex.z-dn.net/?f=cos%282arctg1%29+%3D+cos%282%5Ccdot+%5Cfrac%7B%5Cpi%7D%7B4%7D%29+%3D+cos%5Cfrac%7B%5Cpi%7D%7B2%7D+%3D+0)
в)
![ctg(2arcsin\frac{1}{\sqrt2}) = ctg(2 \underbrace { arcsin\frac{\sqrt2}{2}}_{\frac{\pi}{4}}}) = ctg(2\cdot\frac{\pi}{4}) = ctg\frac{\pi}{2} = 0](https://tex.z-dn.net/?f=ctg%282arcsin%5Cfrac%7B1%7D%7B%5Csqrt2%7D%29+%3D+ctg%282+%5Cunderbrace+%7B+arcsin%5Cfrac%7B%5Csqrt2%7D%7B2%7D%7D_%7B%5Cfrac%7B%5Cpi%7D%7B4%7D%7D%7D%29+%3D+ctg%282%5Ccdot%5Cfrac%7B%5Cpi%7D%7B4%7D%29+%3D+ctg%5Cfrac%7B%5Cpi%7D%7B2%7D+%3D+0)
г)
![cos(\underbrace{arccos(-\frac{1}{2})}_{\pi-arccos(\frac{1}{2})}+\frac{\pi}{3}) = cos(\pi - \frac{\pi}{3}+\frac{\pi}{3}) = cos\pi = -1](https://tex.z-dn.net/?f=cos%28%5Cunderbrace%7Barccos%28-%5Cfrac%7B1%7D%7B2%7D%29%7D_%7B%5Cpi-arccos%28%5Cfrac%7B1%7D%7B2%7D%29%7D%2B%5Cfrac%7B%5Cpi%7D%7B3%7D%29+%3D+cos%28%5Cpi+-+%5Cfrac%7B%5Cpi%7D%7B3%7D%2B%5Cfrac%7B%5Cpi%7D%7B3%7D%29+%3D+cos%5Cpi+%3D+-1)
Вроде так
(8x+112/48)-(6x-72/48)=624/48
8x+112-6x+72=624
8x-6x=624-112-72
2x=440
x=220
Это когда на числовой оси нарисована ПРЯМАЯ. ОТ значения Х зависит ее направление и то, в каких четвертях она лежит