Ответ:
1)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
2)tg(πх/4)=0; πх/4=π ; πх=4π; х=4
tg(πх/12)=1 ;πх/12=π/4; 4πх=12π ; х=12π/4π ; х=3
tg(πх/6)=-1 ;πх/6=-π/4; 4πх=-6π; х=-3/2
tg(πх/3)=√3 ;πх/3=π/3 ;х=1
tg(πх/6)=-√3 ;πх/6=-π/3 ; х=-2
tg(πх/12)=1/√3; πх/12=π/6; х=2
tg(πх/6)=-1/√3 ; πх/6=-π/6 ;х=-1
ctg(πх/4)=0; πх/4=π/2; х=2
ctg(πх/8)=1;πх/8=π/4; х=2
ctg(πх/16)=-1; πх/16=-π/4: х=-4
3) tg(пх/8)=1
πx/8=π/4+πN
x/8=1/4+N
x=2+8N
минимум 2
4)ctg (pix/16) = - 1
pix/16 = 3pi/4 + pik * (16/pi)
x = 12 + 16k
наименьший положительный корень
k= 0 ==> x = 12+16*0 = 12
k = 1 ==> x = 12+16*1 = 28
k = -1 ==> x = 12-16 = - 4
Ответ:12