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Andrey54rus
@Andrey54rus
July 2022
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cos2x+sin4x=sin^2x-cos^2x
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DariosI
Verified answer
Cos2x+sin4x=-(cos^2x-sin^2x)
cos2x+2sin2x*cos2x=-cos2x
cos2x+2sin2x*cos2x+cos2x=0
2cos2x+2sin2x*cos2x=0
2cos2x(1+sin2x/cos2x)=0
2cos2x(1+tg2x)=0
cos2x=0
2x=π/2+π*n
1+tg2x=0
tg2x=-1
2x=-π/4+2π*n
x=-π/8+πn
x=π/4+πn/2
1 votes
Thanks 1
Andrey54rus
квадраты убираются ??
DariosI
Это формула двойного угла cos
Andrey54rus
А, блин, не увидел кое-что
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Answers & Comments
Verified answer
Cos2x+sin4x=-(cos^2x-sin^2x)cos2x+2sin2x*cos2x=-cos2x
cos2x+2sin2x*cos2x+cos2x=0
2cos2x+2sin2x*cos2x=0
2cos2x(1+sin2x/cos2x)=0
2cos2x(1+tg2x)=0
cos2x=0
2x=π/2+π*n
1+tg2x=0
tg2x=-1
2x=-π/4+2π*n
x=-π/8+πn
x=π/4+πn/2