求方程1+2cosx^2+2sin2x=0在[-π,π]上的解集

Question

求方程 1+2cosx^2+2sin2x=0 在[-π,π]上的解集

Answer 1

1+2cosx^2+2sin2x
=2+(2cosx^2-1)+2sin2x
=2+cos2x+2sin2x
=2+√5[(1/√5)cos2x+(2/√5)sin2x]
=2+√5·cos[2x+arccos(1/√5)]
则1+2cosx^2+2sin2x=0 即
2+√5·cos[2x+arccos(1/√5)]=0.
cos[2x+arccos(1/√5)]=-2/√5;
则2x+arccos(1/√5)=arccos(-2/√5)
或=-arccos(-2/√5).
则2x=arccos(-2/√5)-arccos(1/√5)=π-arcsin(1/√5)-arccos(1/√5)
=π-[arcsin(1/√5)-arccos(1/√5)]
=π-π/2
=π/2.
或2x=-arccos(-2/√5)-arccos(1/√5)=-π+arcsin(1/√5)-arccos(1/√5)
=-π+2arcsin(1/√5).