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Нежнее3Ножа
@Нежнее3Ножа
June 2022
1
11
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sinX-√3cosX=2
cosX-sinX=√2
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sedinalana
SinX-√3cosX=2
2(1/2*sinx-
√3/2cosx)=2
sin(x-
π/3)=1
x-
π/3=π/2+2πk
x=5
π/6+2πk,k∈z
cosX-sinX=√2
√2(1/√2*cosx-1/√2*sinx)=√2
cos(x+
π/4)=1
x+π/4=2πk
x=-π/4+2πk,k∈z
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Answers & Comments
2(1/2*sinx-√3/2cosx)=2
sin(x-π/3)=1
x-π/3=π/2+2πk
x=5π/6+2πk,k∈z
cosX-sinX=√2
√2(1/√2*cosx-1/√2*sinx)=√2
cos(x+π/4)=1
x+π/4=2πk
x=-π/4+2πk,k∈z