Помогите с тригонометрией 1+sin^2x=cos^2x+sinxсоs((П/12)+(x/3))-1=0. Created by Fortiorisa. matematika-ru

3))-1=0...

July 2022 1 54 Report

Помогите с тригонометрией
1+sin^2x=cos^2x+sinx
соs((П/12)+(x/3))-1=0

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hgm169899 1+sin²x-cos²x=sinx
1-cos²x=sin²x
sin²x+sin²x=sinx
2sin²x=sinx
2sinx=1
sinx=1/2
x=(-1)^(n)*arcsin1/2+πn
x=(-1)^(n)*π/6+πn

cos((π/12)+(x/3))=1
π/12+x/3=2πn
x/3=-π/12+2πn
x=-π/4+6πn

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