Question

ПОМОГИТЕ ПОЖАЛУЙСТА!!!!!!!!

Answer 1

1.
sin0=0
sin(π/2)=1
sinπ=0
sin(-2π)=0
Ответ: а), в), г)

2.
cos(π/2 +2πn)=0
sin(π/2 +2πn)=1
Ответ: б), г)

3.
а) cos2x=1
2x=2πn
x=πn, n∈Z
б) tg(-x)=-1
tg(x)=1
x=π/4 + πn, n∈Z
в) sin(x-π)=1
sin(π-x)=-1
sinx=-1
x=-π/2 +2πn, n∈Z

4.
a)sin(x+π/6)-1=0
sin(x+π/6)=1
x+π/6=π/2 +2πn
x=π/2+π/6 +2πn=4π/6 +2πn=2π/3 +2πn
x=2π/3 +2πn, n∈Z
б) ctg(π/3+x)=√3
π/3+x=π/6 +πn
x=π/6 - π/3 +πn=- π/6 +πn
x=- π/6 +πn, n∈Z
в) √2 cos(π/4-x)=1
cos(π/4-x)=1/√2
cos(π/4-x)=√2/2
π/4-x=+-π/4+2πn
π/4-x₁=π/4+2πn
-x₁=2πn
x₁=2πn, n∈Z
π/4-x₂=-π/4+2πn
-x₂=-π/2+2πn
x₂=π/2+2πn, n∈Z

5. (cos2x+cos(π/6))(cos(x/2)-2)=0
cos(x/2)-2 никогда не равно 0
поэтому только
cos2x+cosπ/6=0
cos2x=-cosπ/6
cos2x=-√2/2
2x=+-5π/6+2πn
x=+-5π/12+πn
промежутку [0;π] принадлежат x=5π/12 и x=7π/12
сумма π