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julietta18
@julietta18
July 2022
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помогите, как такое решать??
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DariosI
Verified answer
-2cos2x=2-2cosx
-cos2x=1-cosx
-(2cos²x-1)=1-cosx
-2cos²x+1-1+cosx=0
2cos²x-cosx=0
2cosx(cosx-1/2)=0
cosx=0
x=π/2+πn, n∈Z
cosx=1/2
x=+-π/3+2πn n∈Z
ОТВЕТ -3π/2; -7π/3; -5π/3
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Answers & Comments
Verified answer
-2cos2x=2-2cosx
-cos2x=1-cosx
-(2cos²x-1)=1-cosx
-2cos²x+1-1+cosx=0
2cos²x-cosx=0
2cosx(cosx-1/2)=0
cosx=0
x=π/2+πn, n∈Z
cosx=1/2
x=+-π/3+2πn n∈Z
ОТВЕТ -3π/2; -7π/3; -5π/3