1. 2cos2x=7cosx 2. cos^4x-sin^4x=sqrt(3)/2 3. sin^2x-sqrt(3)/2*sin2x=cos^2x - вопрос №1227244323232222192840 от SophiaB 28.05.2021 09:23

Question

Алгебра: 1. 2cos2x=7cosx 2. cos^4x-sin^4x=sqrt(3)/2 3. sin^2x-sqrt(3)/2*sin2x=cos^2x

SophiaB · Answer

1)2(2cos²x-1)-7cosx=04cos²x-7cosx-2=0a=cosx  4a²-7a-2=0 D=49+32=81 √D=9a1=(7-9)/8=-1/4    cosx=-1/4    x=π+ -arccos1/4+2πna2=(7+9)/8=2∉[-1;1]-нет решения2)(cos²x-sin²x)(cos²x+sin²x)=√3/2cos2x=√3/22x=+ - π/6+2πn⇒x=+ - π/12+πn3)cos²x-sin²x+√3/2sin2x=0cos2x-√3/2sin2x=0  /cos2x≠01-√3/2tg2x=0tg2x=2/√3=2√3/32x=arctg2√3/3+πnx=1/2arctg2√3/3+πn/2...