Объяснение:
Уравнение к-ой степени имеет к корней в поле комплексных чисел
cos(π/12)=cos[(π/3)-(π/4)]=cos(π/3)cos(π/4)+sin(π/3)sin(π/4)=
=0,5·(√2/2)+(√3/2)(√2/2)=(√6+√2)/4
sin(π/12)=sin[(π/3)-(π/4)]=sin(π/3)cos(π/4)-cos(π/3)sin(π/4)=
=(√3/2)(√2/2)-0,5·(√2/2)=(√6-√2)/4
x⁶+3i=0
x⁶=-3i=3(cos(-π/2)+isin(-π/2))
x₀= (cos(-π/12)+isin(-π/12))= (cos(π/12)-isin(π/12))=
=((√6+√2)/4-i(√6-√2)/4)=((√6+√2)-i(√6-√2))/4
x₁= (cos((-π+2π)/12)+isin((-π+2π)/12))= (cos((π)/12)+isin((π)/12))=
((√6+√2)+i(√6-√2))/4
x₂= (cos((-π+4π)/12)+isin((-π+4π)/12))= (cos((3π)/12)+isin((3π)/12))=
= (cos((π)/4)+isin((π)/4))= (√2/2+i√2/2)= √2(1+i)/2
x₃= (cos((-π+6π)/12)+isin((-π+6π)/12))= (cos((5π)/12)+isin((5π)/12))=
= (sin((π)/12)+icos((π)/12))= ((√6-√2)/4+i(√6+√2)/4)=
=((√6-√2)+i(√6+√2))/4
x₄= (cos((-π+8π)/12)+isin((-π+8π)/12))= (cos((7π)/12)+isin((7π)/12))=
= (-sin((π)/12)+icos((π)/12))= (-(√6-√2)/4+i(√6+√2)/4)=
=((√2-√6)+i(√6+√2))/4
x₅= (cos((-π+10π)/12)+isin((-π+10π)/12))= (cos((3π)/4)+isin((3π)/4))=
= (-cos((π)/4)+isin((π)/4))= (-√2/2+i√2/2)= √2(-1+i)/2
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1, 3x+0, 2=0, 7x^2 решить с пом дискриминанта