Sinx+2=0 sin2x=-√3/2 2cosx-√2=0 cos5x=0 √3tg2x+1=0 - вопрос №697009820232220503210 от askerova1408 21.09.2020 12:33

Question

Алгебра: Sinx+2=0 sin2x=-√3/2 2cosx-√2=0 cos5x=0 √3tg2x+1=0

askerova1408 · Answer

1) sinx=-2x=(-1)ⁿarcsin(-2)+πn, n∈Z2) sin2x=-√3/2x=(-1)ⁿ arcsin(-√3/2)+πn , n∈Zx=-( (-1)ⁿ  *π/3)+πn , n∈Z3)2cosx=√2cosx=√2/2x=+-arccos(√2/2)+2πn , n∈Zx=+-π/4+2πn , n∈Z4) cos5x=05x=π/2+πn , n∈Zx=π/10+πn/5 , n∈Z5) tg2x=-1/√3x=-arctg(1/√3)+πn , n∈Z2x=-π/6++πn , n∈Zx=-π/12+πn/2 , n∈Z...