2^(sinx)*2^sin3x=16^(cos^2x). Created by pestov2000. algebra-ru

2^(sinx)*2^sin3x=16^(cos^2x)...

July 2022 1 9 Report

2^(sinx)*2^sin3x=16^(cos^2x)

sedinalana

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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