Помогите решить!!! с объяснением, ПЖ!!! 1)2sin (x-П / 3) = корень2 2)cos (3x-П) = -1/2 3)2sin (2П-2x/3) = - корень из 2

2sin(x-π/3)=√2
sin(x-π/3)=√2/2
x-π/3=(-1)^n *arcsin(√2/2)+πn, n∈Z
x-π/3=(-1)^n *(π/4)+πn, n∈Z
<u>x=(-1)^n*(π/4)+π/3+πn, n∈Z</u>

cos(3x-π)=-1/2
cos(-(π-3x))=-1/2
cos(π-3x)=-1/2  [cos(-α)=cosα,  четная]
-cos3x=-1/2, cos3x=1/2
3x=+-arccos(1/2)+2πn, n∈Z
3x=+-π/3+2πn, n∈Z  |:3
<u>x=+-π/9+(2/3)*πn, n∈Z</u>

2sin(2π-(2/3)x)=-√2
sin(2π-(2/3)x)=-√2/2
-sin((2/3)x)=-√2/2,   sin((2/3)x)=√2/2
(2/3)x=(-1)^n *arcsin(√2/2)+πn, n∈Z
(2/3)x=(-1)^n *(π/4)+πn, n∈Z  |:2/3
<u>x=(-1)^n *(π/6)+(3/2)πn, n∈Z</u>