Найти корни

sinx=1/2 x принадлежит [-2p; 2p]

sinx=√2/2 x принадлежит [-p/2; 2p]

sinx-√2/2=0 x принадлежит [-2p; p/2]

1) sinx=1/2; x∈[-2π; 2π]

x = (-1) ⁿπ/6+πn

x=-11π/6; -7π/6; π/6; 5π/6∈[-2π; 2π]

2) sinx=√2/2; x∈[-π; 2π]

x = (-1) ⁿπ/4+πn

x=π/4; 3π/4∈[-π; 2π]

3) sinx=√2/2; x∈[-2π; π/2]

x = (-1) ⁿπ/4+πn

x=-7π/4; -5π/4; π/4∈[-2π; π/2]