Ответы на вопрос:
Sin(2x-2)=2sin(x-1)*cos(x-1) sin2x-six2=2sin(x-1)*cos(x+1) 2sin(x-1)*cos(x-1)=2sin(x-1)*cos(x+1) sin(x-1)*cos(x-1)-sin(x-1)*cos(x+1)=0 sin(x-1)*(cos(x-1)-cos(x+1))=0 sin(x-1)=0; cos(x-1)-cos(x+1)=0 x-1=pi*n ; cosx*cos1+sinx*sin1-cosx*cos1+sinx*sin1=0 x=1+pi*n; 2sinx*sin1=0 sinx=0 x=pi*k [0; 2*pi] [0; 2*pi] x=1+pi*n x=pi*k n=0, x=1. k=0, x=0 n=1, x=1+pi k=1, x=pi, k=2, x=2*pi ответ: 0, 1, pi, 1+pi, 2pi,
Популярно: Математика
-
Cookiefluffy150018.06.2021 08:34
-
popovapp2620.06.2022 17:58
-
SeetFatRed81912.08.2022 07:44
-
Audana0010.10.2022 06:19
-
alenka1812106.03.2022 19:23
-
geraveselov2017.08.2022 03:45
-
2006zahar200626.02.2020 20:55
-
geracim127.05.2020 15:22
-
sofa456578712.10.2022 11:48
-
макс1071025.11.2021 12:40