Ответы на вопрос:
1).cos4x=-1 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)=0воспользуемся 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
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