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TatyanaBergman
@TatyanaBergman
August 2022
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Решите уравнение:
а) sin²x+sinx-2=0;
б) 3sin²x-cosx+1=0.
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mikael2
а) sin²x+sinx-2=0 sinx=z z²+z-2=0 z1= -2 z2=1
-2 не подходит sinx=1 x=π/2+2πn n∈Z
б) 3sin²x-cosx+1=0. 3(1-cos²x)-cosx+1=0
-3cos²x-cosx+4=0 3cos²x+cosx-4=0 √ D=√1+48=7
x1=1/6[-1-7]<-1 не подходит
x2=1/6[-1+7]=1 cosx=1 x=0+2πn=2πn n∈Z
2 votes
Thanks 2
TatyanaBergman
Спасибо большое!
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Answers & Comments
-2 не подходит sinx=1 x=π/2+2πn n∈Z
б) 3sin²x-cosx+1=0. 3(1-cos²x)-cosx+1=0
-3cos²x-cosx+4=0 3cos²x+cosx-4=0 √ D=√1+48=7
x1=1/6[-1-7]<-1 не подходит
x2=1/6[-1+7]=1 cosx=1 x=0+2πn=2πn n∈Z