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]