Решите уравнении:
1. cos4x=-1
2. sin(4x-π/3)=1/2
3.2sin^2(x/2)=1
4. cos4x-cos5x=0
5.cosx- корень из cosx =0
6. cos2x*cos(x+π/6)+sin2x*sin(x+π/6)=0
С решениями пж
1. cos4x=-1
2. sin(4x-π/3)=1/2
3.2sin^2(x/2)=1
4. cos4x-cos5x=0
5.cosx- корень из cosx =0
6. cos2x*cos(x+π/6)+sin2x*sin(x+π/6)=0
С решениями пж
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Answers & Comments
4x=π+2πn, x=π/4+πn/2,n∈Z,
2). sin(4x-π/3)=1/2
4x-π/3=(-1)ⁿπ/6+πn, n∈Z, 4x=π/3+(-1)ⁿπ/6+πn ,
x= 4π/3+ (-1)ⁿ 2π/3+4πn , n∈Z.
3).2sin²(x/2)=1, sin²(x/2)=1/2, sin(x/2)=1/√2, sin(x/2)=-1/√2
x/2=(-1)ⁿπ/4+πn,n∈Z, x/2=(-1)ⁿ⁺¹/π4+ πn, n∈Z
x=(-1)ⁿπ/2+2πn,n∈Z, x= (-1)ⁿπ/2+2πn,n∈Z,
4). cos4x-cos5x=0
cos4x-cos5x=-2sin(4x+5x)/2·sin(4x-5x)/2=0
-2sin(4,5x)·sin(-0,5x) =2sin 4,5x·sin0,5x=0, sin 4,5x·sin0,5x=0
sin4,5x=0, sin0,5x=0
4,5x=πn 0,5x=πn
9x/2=πn x/2=πn/ n∈Z
x=2πn/9 x=2πn, n∈Z
5.cosx- √cosx =0, √cosx(√cosx-1)=0
√cosx=0 √cosx=1
cosx=0 cosx=1
x=π/2 +2πn , n∈Z x=2πn , n∈Z
6. cos2x*cos(x+π/6)+sin2x*sin(x+π/6)=0
Воспользуемся формулой:
cosαcosβ+sinαsinβ=cos(α-β)
cos2x*cos(x+π/6)+sin2x*sin(x+π/6)=cos(2x-(x+π/6))=cos(2x-x-π/6)=0
cos(x-π/6)=0, x-π/6=π/2+2πn, x=π/6+π/2+2πn,n∈Z
x=(π+3π)/6+2πn,n∈Z, x=4π/6++2πn,n∈Z, x=2π/3+2πn.n∈Z