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5рома5
@5рома5
July 2022
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Помогите решить уравнение:
sin3x-sinx+cos2x=1
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sedinalana
Verified answer
Sin3x-sinx+cos2x=1
2sinxcos2x+cos²x-sin²x-sin²x-cos²x=0
2sinxcos2x-2sin²x=0
2sinx*(cos2x-sinx)=0
sinx=0⇒x=πn,n∈z
cos2x-sinx=0
1-2sin²x-sinx=0
sinx=a
2a²+a-1=0
D=1+8=9
a1=(-1-3)/4=-1⇒sinx=-1⇒x=-π/2+2πn,n∈z
a2=(-1+3)/4=1/2⇒sinx=1/2⇒x=(-1)^n*π/6+πn,n∈z
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Answers & Comments
Verified answer
Sin3x-sinx+cos2x=12sinxcos2x+cos²x-sin²x-sin²x-cos²x=0
2sinxcos2x-2sin²x=0
2sinx*(cos2x-sinx)=0
sinx=0⇒x=πn,n∈z
cos2x-sinx=0
1-2sin²x-sinx=0
sinx=a
2a²+a-1=0
D=1+8=9
a1=(-1-3)/4=-1⇒sinx=-1⇒x=-π/2+2πn,n∈z
a2=(-1+3)/4=1/2⇒sinx=1/2⇒x=(-1)^n*π/6+πn,n∈z