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]