Г
(1-2sin²(x/2))²=cos²x
2tg^4(x/2)/(1+tg²(x/2))=2sin^4(x/2)/cos^4(x/2):1/cos^4(x/2)=2sin^4(x/2)=
=2*(1-cosx)²/4=(1-2cosx+cos²x)/2
Получаем
cos²x-(1-2cosx+cos²x)/2=√3/2
2cos²x-1+2cosx-cos²x=√3
cos²x+2cosx-(1+√3)=0
cosx=a
a²+2a-(1+√3)=0
D=4+4+4√3=8+4√3=4(2+√3)
a1=(-2-2√*2+√3)/2=-1-√(2+√3)⇒cosx=-1-√(2+√3)<-1 нет решения<br>a2=-1+√(2+√3)⇒cosx=-1+√(2+√3)⇒x=+-arccos(√(2+√3)-1)+2πn,n∈z
ж
(1+cos2x)/2+(1+cos4x)/2+(1+cos6x)/2+(1+cos8x)/2=2
4+(cos2x+cos8x)+(cos4x+cos6x)=4
2cos5xcos3x+2cos5xcosx=0
2cos5x(cos3x+cosx)=0
4cos5xcos2xcosx=0
cos5x=0⇒5x=π/2+πn,n∈z⇒x=π/10+πn/5,n∈z
cos2x=0⇒2x=π/2+πk,k∈z⇒x=π/4+πk/2,k∈z
cosx=0⇒x=π/2+πm,m∈z
д
1/2*(cos5x+cos8x)=1/2*(cos2x+cos9x)
cos5x=cos2x
cos5x-cos2x=0
2sin(3x/2)*sin(7x/2)=0
sin(3x/2)=0⇒3x/2=πn⇒x=2πn/3,n∈z
sin(7x/2)=0⇒7x/2=πk,k∈z⇒x=2πk/7,k∈z