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