Ответ:
-π/15+πk/3, -π/8 ±3π/8 + πk, k∈Z, (-1)^kπ/2 +3πk/2, ±2π/3 +2πk, k∈Z
Пошаговое объяснение:
1) tg(π/5+3x) = 0
π/5+3x=arctg0+πk, k∈Z
π/5+3x= πk
3x= -π/5 +πk
x= -π/15+πk/3, k∈Z
2) cos(π/4+2x)= -√2/2
π/4+2x=±arccos(-√2/2)+2πk, k∈Z
2x=-π/4 ±3π/4+2πk
x= - π/8 ±3π/8+πk, k∈Z
3) 4sin(x/3)cos(x/3) = 2√3
2sin(2x/3) = √3
sin(2x/3)=√3/2
2x/3=( -1)^k arcsin(√3/2)+πk
2x/3 = (-1)^k π/3 + πk
x= ( -1)^k π/2 + 3πk/2, k∈Z
4) D=(-7)²- 4*2*(-4)=81
cosx=(7±9)/4
cosx =4, решений нет;
cosx= -1/2
x= ±arccos(-1/2) +2πn
x= ± 2π/3+2πn, n∈Z