Ответ:
2cos(3x-π/3)-√3=0
2cos(3x-π/3)=√3
cos(3x-π/3)=√3/2
3x-π/3=+/-π/6+2πn, n∈Z
3x=+/-π/6+π/3+2πn, n∈Z
x=+/-π/18+π/9+2/3πn, n∈Z
x1=π/18+π/9+2/3πn=π/6+2/3πn, n∈Z
x2 =-π/18+π/9+2/3πn=π/18+2/3πn, n∈Z
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Ответ:
2cos(3x-π/3)-√3=0
2cos(3x-π/3)=√3
cos(3x-π/3)=√3/2
3x-π/3=+/-π/6+2πn, n∈Z
3x=+/-π/6+π/3+2πn, n∈Z
x=+/-π/18+π/9+2/3πn, n∈Z
x1=π/18+π/9+2/3πn=π/6+2/3πn, n∈Z
x2 =-π/18+π/9+2/3πn=π/18+2/3πn, n∈Z