A maths problem

4(cosx-sinx)+ksinx=5
4cosx=5+(4-k)sinx
4(1-(sinx)^2)^1/2=5+(4-k)sinx
16(1-(sinx)^2)=25+10(4-k)sinx+(4-k)^2(sinx)^2
(k^2-8k+32)(sinx)^2+10(4-k)sinx+9=0
the eq. has no sol.
so delta<0
(10(4-k))^2-4(9)(k^2-8k+32)<0
k^2-8k+7<0
1<7

2.(secx)^2(1-sinx)=2-cosx-sinx
(1+(tanx)^2)(1-sinx)=2-cosx-sinx
((cosx)^2+(sinx)^2)(1-sinx)=2(cosx)^2-(cosx )^3-(cosx)^2sinx
(cosx)^2-(sinx)^2=(cosx)^3-(sinx)^3
(cosx-sinx)(cosx+sinx-1+cosxsinx)=0
cosx=sinx or cosx+sinx-1+cosxsinx=0
tanx=0 or cosx+sinx=1-cosxsinx
x=0,p,2p or 1+2cosxsinx=1-2cosxsinx+(cosxsinx)^2
1+sin2x=1-sin2x+(sin2x/2)^2
8sin2x=(sin2x)^2
sin2x(sin2x-8)=0
sin2x=0 or sin2x=8(rejected)
x=0,p/2,p,3p/2,2p

1.
4(cosx -sinx) +ksinx =5
(k-4)sinx+4cosx=5
√(k²-8k+20)sin[x+tan-14/(k-4)]=5
sin[x+tan-14/(k-4)]=5/ √(k²-8k+20)
=>5/ √(k²-8k+20)>1or5/ √(k²-8k+20)<-1
5>√(k²-8k+20) or -5<√(k²-8k+20) (rejected)
25>k²-8k+20
k²-8k-5<0
4-√21<k<4+√21

2.
sec²x=(2-cosx-sinx)/(1-sinx)
(1+tan²x)(1-sinx)=2-cosx-sinx
1+tan²x-sinx-sinxtan²x=2-cosx-sinx
sin²x-sin³x=cos²x-cos³x
sin²x-cos²x=sin³x-cos³x
(sinx+cosx)(sinx-cosx)=(sinx-cosx)(sin²x+sinxcosx+cos²x)
(sinx-cosx)(sinx+cosx-1-sinxcosx)=0
(sinx-cosx)(1-sinx)(cosx-1)=0
sinx=cosx or sinx=1 or cos x=1
tanx=1 or x=2npi+pi/2 or2npi
x=npi+pi/4 or 2npi+pi/2 or2npi




2007-12-28 13:15:37 補充：
修正:sinx=1(rejected)

2007-12-28 20:02:11 補充：
修正:1.4(cosx -sinx) +ksinx =5(k-4)sinx+4cosx=5√(k²-8k+32)sin[x+tan^-14/(k-4)]=5sin[x+tan-14/(k-4)]=5/ √(k²-8k+32)=&gt;5/ √(k²-8k+32)&gt;1or5/ √(k²-8k+32)&lt;-15&gt;√(k²-8k+32) or -5&lt;√(k²-8k+32 (rejected) 25&gt;k²-8k+32k²-8k+7&lt;0(k-1)(k-7)=01&lt;7

2007-12-28 20:03:16 補充：
2.sec²x=(2-cosx-sinx)/(1-sinx)(1+tan²x)(1-sinx)=2-cosx-sinx1+tan²x-sinx-sinxtan²x=2-cosx-sinxsin²x-sin³x=cos²x-cos³xsin²x-cos²x=sin³x-cos³x

2007-12-28 20:03:27 補充：
(sinx+cosx)(sinx-cosx)=(sinx-cosx)(sin²x+sinxcosx+cos²x)(sinx-cosx)(sinx+cosx-1-sinxcosx)=0(sinx-cosx)(1-sinx)(cosx-1)=0sinx=cosx or sinx=1(rejected)or cos x=1tanx=1 or x=0,360x=0,45,225,360