Нахождение корней

sin=1/2 x принадлежит[-3p; 0]

sinx-1/2=0 x принадлежит[p/2; 5p/2]

Sinx=1/2

x=π/6+2πn, n∈z U x=5π/6+2πk, k∈z

a) - 3π≤π/6+2πn≤0

-18≤1+12n≤0

-19≤12n≤-1

-19/12≤n≤-1/12

n=-1⇒x=π/6-2π=-11π/6∈[-3π; 0]

-3π≤5π/6+2πk≤0

-18≤5+12k≤0

-23≤12k≤-5

-23/12≤k≤-5/12

k=-1⇒x=5π/6-2π=--7π/6∈[-3; 0]

b) π/2≤π/6+2πn≤5π/2

3≤1+12n≤15

2≤12n≤14

1/6≤n≤1 1/6

n=1⇒x=π/6+2π=13π/6∈[π/2; 5π/2]

π/2≤5π/6+2πk≤5π/2

3≤5+12k≤15

-2≤12k≤10

-1/6≤k≤5/6

k=0⇒x=5π/6∈[π/2; 5π/2]