Question

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]

Answer 1

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]