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juliiasemibratova13
@juliiasemibratova13
July 2022
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ПОМОГИТЕ РЕШИТЬ, СРОЧНО!!!! ПОЖАЛУЙСТААААААААААААААААА!!!!!!!!!!!!
1. cos2x+sin²x=cosx+2
2. cos²x-cos2x=2-sinx
3. 3sin2x+8cos²x=7
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basilionin
1. c
os2x+sin²x=cosx+2
2cos
²x-1+1-cos²x=cosx+2
cos²x-cosx-2=0
Делаем замену t=cosx, по модулю t не превосходит 1
t²-t-2=0
t = -1 или 2 (не подходит)
cosx = -1
x = pi + 2pin
2.
cos²x-cos2x=2-sinx
sin²x+sinx=2
Аналогично, sinx=1
x=pi/2+2pin
2 votes
Thanks 1
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Answers & Comments
2cos²x-1+1-cos²x=cosx+2
cos²x-cosx-2=0
Делаем замену t=cosx, по модулю t не превосходит 1
t²-t-2=0
t = -1 или 2 (не подходит)
cosx = -1
x = pi + 2pin
2. cos²x-cos2x=2-sinx
sin²x+sinx=2
Аналогично, sinx=1
x=pi/2+2pin