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Саныч89
@Саныч89
June 2022
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sinx+cosx=sinxcosx+1
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Sinx+cosx=sinxcosx+1 ;
2(Sinx+cosx)= 2sinxcosx+2 ;
2(Sinx+cosx)=(sinx+cosx)²+1;
(sinx+cosx)²-2(Sinx+cosx)+1 =0 ;
(sinx +cosx -1)² =0;
sinx +cosx =1 ; * * * 1/√2sinx +1/√2cosx =1/√2 ⇔(cosx*cosπ/4+sinx*sinπ/4) = 1/√2 * * *
cos(x -π/4) =1/√2 ;
x -π/4 =± π/4+2πn , n∈Z.
x = π/4 ± π/4+2πn , n∈Z.
x₁ =2πn ; x₂ = π/2 +2πn , n∈Z.
ответ: 2πn ; π/2 +2πn , n∈Z.
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Verified answer
Sinx+cosx=sinxcosx+1 ;2(Sinx+cosx)= 2sinxcosx+2 ;
2(Sinx+cosx)=(sinx+cosx)²+1;
(sinx+cosx)²-2(Sinx+cosx)+1 =0 ;
(sinx +cosx -1)² =0;
sinx +cosx =1 ; * * * 1/√2sinx +1/√2cosx =1/√2 ⇔(cosx*cosπ/4+sinx*sinπ/4) = 1/√2 * * *
cos(x -π/4) =1/√2 ;
x -π/4 =± π/4+2πn , n∈Z.
x = π/4 ± π/4+2πn , n∈Z.
x₁ =2πn ; x₂ = π/2 +2πn , n∈Z.
ответ: 2πn ; π/2 +2πn , n∈Z.