Решите уравнение cos(2x-п\3)=-1 2 cos3x=корень из 3 2 sin(x\2-п\6)=-1

Cos(2x-p/3)=1
T=2x-p/3
Cost=1
T=2pk
2x-p/3=2pk
2x=p/3+2pk
X=p/6+pk

2cos3x=sqrt3
T=3x
2cost=sqrt3
Cost=(sqrt3)/2
T=p/6+2pk;-p/6+2pk
3x=p/6+2pk
X=p/9+(2pk)/3
3x=-p/6+2pk
X=-p/9+(2pk)/3

2sin(x/2-p/6)=-1
T=x/2-p/6
2sint=-1
Sint=-1/2
T=((-1)^(k+1))*p/6+pk
X/2-p/6=((-1)^(k+1))*p/6+pk
X/2=((-1)^(k+1))*p/6+p/6+pk
X=((-1)^(k+1))*p/3+p/3+2pk