2x^2+5x-3=0
метод переброски
x^2+5x-6=0
по теореме Виета
x1=-3/2 x2=-2/2
x1=-1,5 x2=-1
по формуле раскладываем на множители
(x+1,5)*(x+1)=0
![a) \ x_k= \frac{ \pi }{2} + \pi k, k \in Z, \\ y_k= \frac{ \pi }{3} +2 \pi k, k \in Z,\\ z_k=\frac{2 \pi }{3} +2 \pi k, k \in Z\\](https://tex.z-dn.net/?f=a%29++%5C+x_k%3D+%5Cfrac%7B+%5Cpi+%7D%7B2%7D+%2B+%5Cpi+k%2C+k+%5Cin+Z%2C+%5C%5C%0A+y_k%3D++%5Cfrac%7B+%5Cpi+%7D%7B3%7D+%2B2+%5Cpi+k%2C+k+%5Cin+Z%2C%5C%5C%0A+z_k%3D%5Cfrac%7B2+%5Cpi+%7D%7B3%7D+%2B2++%5Cpi+k%2C+k+%5Cin+Z%5C%5C+)
b) С помощью числовой окружности отберем корни, принадлежащие отрезку
![[- \frac{5 \pi }{2} ;- \pi ]](https://tex.z-dn.net/?f=%5B-+%5Cfrac%7B5+%5Cpi+%7D%7B2%7D+%3B-+%5Cpi+%5D)
.
Получаем числа:
![x_{-2}= \frac{ \pi }{2} -2 \pi =- \frac{3 \pi }{2}\\ y_{-1}= \frac{ \pi }{3} -2 \pi=- \frac{5 \pi }{3} \\ z_{-1}= \frac{2 \pi}{3} -2 \pi =- \frac{4 \pi }{3} ](https://tex.z-dn.net/?f=x_%7B-2%7D%3D+%5Cfrac%7B+%5Cpi+%7D%7B2%7D+-2+%5Cpi+%3D-+%5Cfrac%7B3+%5Cpi+%7D%7B2%7D%5C%5C%0Ay_%7B-1%7D%3D+%5Cfrac%7B+%5Cpi++%7D%7B3%7D+-2+%5Cpi%3D-+%5Cfrac%7B5+%5Cpi++%7D%7B3%7D+%5C%5C%0Az_%7B-1%7D%3D+%5Cfrac%7B2+%5Cpi%7D%7B3%7D+-2+%5Cpi+++%3D-+%5Cfrac%7B4+%5Cpi+%7D%7B3%7D+%0A%0A)
Ответ:
![b) \ - \frac{4 \pi }{3}, - \frac{3 \pi }{2}, - \frac{5 \pi }{3} .](https://tex.z-dn.net/?f=b%29+%5C+-+%5Cfrac%7B4+%5Cpi+%7D%7B3%7D%2C+-+%5Cfrac%7B3+%5Cpi+%7D%7B2%7D%2C+-+%5Cfrac%7B5+%5Cpi+%7D%7B3%7D+.)
Sinx = 1
x = π/2 + 2πk
0 ≤ π/2 + 2πk ≤ 4π
-π/2 ≤ 2πk ≤ 7π/2
-1/4 ≤ k ≤ 7/4
k = 0; 1
x = π/2
x = π/2 + 2π = 5π/2
P=M*0,15
M=N*0,18
N=10000*0,40=4000
P=N*0,18*0,15=N*0,027
P=4000*0,027=108