tan{θ+(π/4)}=-√3
θ+(π/4)=xとすると、
tan(x)=-√3
x={(2π)/3}+kπ
(k∈Z kは整数)
θ+(π/4)=x
θ=x-(π/4)
={(2π)/3}+kπ-π/4
={(8π)/12}+kπ-3π/12
={(5π)/12}+kπ //
(k∈Z)
tan{θ+(π/4)}=-√3
θ+(π/4)=xとすると、
tan(x)=-√3
x={(2π)/3}+kπ
(k∈Z kは整数)
θ+(π/4)=x
θ=x-(π/4)
={(2π)/3}+kπ-π/4
={(8π)/12}+kπ-3π/12
={(5π)/12}+kπ //
(k∈Z)
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