Cos(2arccos1/3) = 2cos^2(arccos1/3) - 1 (по формуле двойного аргумента)
Πх/6 = arctg√3 + πk , k ∈ Z
πx/6 = π/3 + πk , k ∈ Z
x/6 = 1/3 + k , k∈Z
x = 2 + 2k , k ∈Z
x = -2
A)2sinπ/6cos(3π/2+x)=2*1/2*sinx=sinx
б)2cos(π+x)cosπ/3=2*1/2*(-cosx)=-cosx
в)(-2sinasinπ/4)/-√2sina=√2sina/√2sina=1
г)√3cosa/(2sinπ/3cosa)=√3:(2*√3/2)=√3/√3=1
д)(cosx-cos3x)/(sin²x+cos²x+cos²x-sin²x)=(-2sin(-x)sin2x)/2cos²x=
4sin²xcosx/2cos²x=2tgxsinx
e)(sin6x+sin2x)/(sin²2x+cos²2x+cos²2x-sin²2x)=2sin4xcos2x/2cos²2x=sin4x/cos2x=
2sin2xcos2x/cos2x=2sin2x