cos x=t,
6t²-7t-5=0,
D=169,
t1=-½, t2=1 ⅔>1,
cos x=-½,
x=±arccos (-½) + 2πk, k∈Z,
x=±(π-arccos (½)) + 2πk, k∈Z,
x=±(π-π/3) + 2πk, k∈Z,
x=±2π/3 + 2πk, k∈Z,
-7π/2<±2π/3 + 2πk<-5π/2,
[-7π/2-2π/3<2πk<-5π/2-2π/3, -7π/2+2π/3<2πk<-5π/2+2π/3,
[-25π/6<2πk<-19π/6, -17π/6<2πk<-11π/6,
[-25/12<k<-19/12, -17/12<k<-11/12,
[k=-2, k=-1,
x=2π/3 - 4π,
x=-10π/3;
x=-2π/3-2π,
x=-8π/3.
Вот пиши ,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,
6х-3<-17-(-x-5)
6x-3<-17+x+5
5x<-17+5+3
5x<-9
x<-9/5
Ответ: (-∞;-9/5)
Только 2:
F=ma
При f=84 и а=12
m12=84
m=7
Ответ:m=7.
+ - +
___________0_________√5__________
y∈(0;√5)
0 < x-3 < √5
3 < x < 3+√5
<u>
x∈(3;3+√5)</u>