2-3cosx-2=0 cos3x+cosx=0 sinx - cosx≥0 tgx≤ - вопрос №82837502320300 от suvaeva884p019j5 15.04.2022 01:07

Question

Математика: 2-3cosx-2=0 cos3x+cosx=0 sinx - cosx≥0 tgx≤

suvaeva884p019j5 · Answer

2cos²x-3cosx-2=0   D=9+16=25  √25=5cosx1=1/4[3+5]=2 не подходитcosx=1/4[3-5]=-1/2x=+-2π/3+2πk   k∈Zcos3x+cosx=02cos2xcosx=0cos2x=0   2x=π/2+2πk   x=π/4+πk   k∈Zcosx=0  x=π/2+2πksinx -√3/3       (π-arcsin(-√3/3))(-1)^k+2πk x arcsin(-√3/3)(-1)^k+2πkk∈Zcosx≥0       -π/2+2πk ≤x≤π/2+2πk   k∈Ztgx≤√3         -π/2+πk x≤π/3+πk   k∈Z...

2-3cosx-2=0 cos3x+cosx=0 sinx> - cosx≥0 tgx≤