Пожалуйста, помогите с тригонометрией, 10 класс
sin²x/4 - cos²x/4 = 1
sin2xcos2x=-1/4
sin3xcosx-cos3xsinx=√3/2
sin x/3cos π/5 - cos x/3sin π/5=√2/2
sin²x-sin2x=0
sin²x/4 - cos²x/4 = 1
sin2xcos2x=-1/4
sin3xcosx-cos3xsinx=√3/2
sin x/3cos π/5 - cos x/3sin π/5=√2/2
sin²x-sin2x=0
More Questions From This User See All
Answers & Comments
-(cos²x/4-sin²x/4)=1
cosx/2=-1
x/2=π+2πn
x=2π+4πn n∈Z
sin2xcos2x=-1/4
2sin2xcos2x/2=-1/4
sin4x=-1/2
4x=(-1)^n×arcsin(-1/2)+πn arcsin(-1/2)=-π/6
4x=(-1)^n×(-π/6)+πn
x=(=1)^n+1×π/24+π/4n n∈Z
sin3xcosx-cos3xsinx=√3/2
sin(3x-x)=√3/2
sin2x=√3/2
2x=(-1)^n×arcsin√3/2+πn arcsin√3/2=π/3
2x=(-1)^n×π/3+πn
x=(-1)^n×π/6+πn/2 n∈Z
sinx/3cosπ/5-cosx/3sinπ/5=√2/2
sin(x/3-π/5)=√2/2
x/3-π/5=(-1)^n×arcsin√2/2+πn arcsin√2/2=π/4
x/3-π/5=(-1)^n×π/4+πn
x/3=(-1)^n×π/4+π/5+πn
x=(-1)^n×3π/4+3π/5+3πn n∈Z
sin²x-sin2x=0
sin²x-2sinxcox=0
sinx(sinx-2cosx)=0
sinx=0 sinx-2cosx=0 |:cosx
x=πn tgx-2=0
tgx=2
x=arctg2+πn n∈Z