4條中4amths數-口-高手救我呀=.=超難呀

2008-02-12 8:15 am
solve the following equations for 0度<=x<=360度.

1.sin2x-∫3cosx=0

2.cos2x=sin^4x+cos^4x

3.2sin^2x-cos2x+1=0

4.2sin^2x-sin2x-2=0

ps:請答得詳盡少少-w-

唔該!

回答 (1)

2008-02-12 10:32 pm
✔ 最佳答案
1. sin2x-∫3cosx=0
2sinxcosx-√3cosx=0
cosx(2sinx-√3)=0
cosx=0 ______________ 2sinx-√3=0
x=90,270 ______________ sinx=√3/2
______________________ x=60,120
_


2. cos2x=sin^4x+cos^4x
cos^2x-sin^2x=sin^4x+cos^4x
cos^2x- cos^4x= sin^4x+ sin^2x
cos^2x(1- cos^2x)= sin^4x+ sin^2x
(1-sin^2x)sin^2x= sin^4x+ sin^2x
sin^2x- sin^4x= sin^4x+ sin^2x
0=2 sin^4x
sin^4x=0
sin^2x=0
x=0,180,360



3. 2 sin^2x -cos2x+1=0
2 sin^2x-(1- 2 sin^2x)+1=0
4 sin^2x-1+1=0
4 sin^2x=0
sin^2x=0
sinx=0
x=0,180,360




4. 2 sin^2x -sin2x-2=0
2 sin^2x-2sinxcosx-2=0
sin^2x-sinxcosx-1=0
sin^2x-sinxcosx- sin^2x-cos^2x=0
cos^2x+sinxcosx=0
cosx(cosx+sinx)=0
cosx=0
x=90,270 _________________cosx+sinx=0
________________ sinx=-cosx
_________________ sinx/cosx =-1
___________________ tanx=-1
_________________ x=135,315
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