sin(x+派/3)sin(x+11派/6)=1/2定义域是 :负派 到 派/2
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sin(x+派/3)sin(x+11派/6)=1/2定义域是 :负派 到 派/2
sin(x+派/3)sin(x+11派/6)=1/2
定义域是 :负派 到 派/2
摆脱了~
sin(x+派/3)sin(x+11派/6)=1/2
定义域是 :负派 到 派/2
摆脱了~
sin(x+π/3)sin(x+11π/6)=1/2
则有:
sin(x+π/3)sin[(x-π/6)+2π]=1/2
sin(x+π/3)sin(x-π/6)=1/2 -----(1)
由于:
sin(x+π/3)
=sinxcos(π/3)+sin(π/3)cosx
=(1/2)sinx+(√3/2)cosx
sin(x-π/6)
=sinxcos(π/6)-sin(π/6)cosx
=(√3/2)sinx-(1/2)cosx
则代入(1)得:
[(1/2)sinx+(√3/2)cosx]*(√3/2)sinx-(1/2)cosx=1/2
(√3/4)(sinx)^2-(√3/4)(cosx)^2+(1/2)sinxcosx=1/2
(-√3/4)[(cosx)^2-(sinx)^2]+(1/4)(2sinxcosx)=1/2
(-√3/4)cos2x+(1/4)sin2x=1/2
(1/2)sin2x-(√3/2)cos2x=1
(cosπ/3)sin2x-(sinπ/3)cos2x=1
sin(2x-π/3)=1
由于:定义域为x属于[-π,π/2]
则:2x-π/3属于[-7π/3,2π/3]
则可得:2x-π/3=π/2或-3π/2
则;x=5π/12 或 -7π/12
则有:
sin(x+π/3)sin[(x-π/6)+2π]=1/2
sin(x+π/3)sin(x-π/6)=1/2 -----(1)
由于:
sin(x+π/3)
=sinxcos(π/3)+sin(π/3)cosx
=(1/2)sinx+(√3/2)cosx
sin(x-π/6)
=sinxcos(π/6)-sin(π/6)cosx
=(√3/2)sinx-(1/2)cosx
则代入(1)得:
[(1/2)sinx+(√3/2)cosx]*(√3/2)sinx-(1/2)cosx=1/2
(√3/4)(sinx)^2-(√3/4)(cosx)^2+(1/2)sinxcosx=1/2
(-√3/4)[(cosx)^2-(sinx)^2]+(1/4)(2sinxcosx)=1/2
(-√3/4)cos2x+(1/4)sin2x=1/2
(1/2)sin2x-(√3/2)cos2x=1
(cosπ/3)sin2x-(sinπ/3)cos2x=1
sin(2x-π/3)=1
由于:定义域为x属于[-π,π/2]
则:2x-π/3属于[-7π/3,2π/3]
则可得:2x-π/3=π/2或-3π/2
则;x=5π/12 或 -7π/12
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