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pestov2000
@pestov2000
July 2022
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2^(sinx)*2^sin3x=16^(cos^2x)
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sedinalana
Verified answer
2^(sinx+sin3x)=2^4cos2x
sinx+sin3x=4cos²x
2sin2xcosx-4cos²x=0
4sinxcos²x-4cos²x=0
4cos²x(sinx-1)=0
cosx=0⇒x=π/2+πk,k∈z
sinx=1πx=π/2+πk,k∈z
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Answers & Comments
Verified answer
2^(sinx+sin3x)=2^4cos2xsinx+sin3x=4cos²x
2sin2xcosx-4cos²x=0
4sinxcos²x-4cos²x=0
4cos²x(sinx-1)=0
cosx=0⇒x=π/2+πk,k∈z
sinx=1πx=π/2+πk,k∈z