tg(пи/5 + 3x)=0
2cos(пи/4 + 2x)= - корень из 2
4sin x/3 * cos x/3 = 2 корень из 3
2cos^2x - 7cosx - 4 = 0

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