Question

1.  2cos2x=7cosx

2.  cos^4x-sin^4x=sqrt(3)/2

3.  sin^2x-sqrt(3)/2*sin2x=cos^2x

 

 

 

Answer 1

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