已知cos(2x+π/3)=-1/2,且x∈【-π/6,π/3】,求角x.

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已知cos(2x+π/3)=-1/2,且x∈【-π/6,π/3】,求角x.已知cos(2x+π/3)=-1/2,且x∈【-π/6,π/3】,求角x.已知cos(2x+π/3)=-1/2,且x∈【-π/

已知cos(2x+π/3)=-1/2,且x∈【-π/6,π/3】,求角x.
已知cos(2x+π/3)=-1/2,且x∈【-π/6,π/3】,求角x.

已知cos(2x+π/3)=-1/2,且x∈【-π/6,π/3】,求角x.
x∈【-π/6,π/3】
2x∈【-π/3,2π/3】
2x+π/3∈【0,π】
所以

cos(2x+π/3)=-1/2

2x+π/3=2π/3
2x=π/3
x=π/6

2x+π/3=2kπ+2π/3
2x=2kπ+π/3
x=kπ+π/6
又x∈【-π/6,π/3】
所以k=0
所以x=π/6