a.maths

a.maths
solve for 0<=x<3 60

(1-tanx)(1+sin2x)=1+ tanx

解：　 　 ( 1 - tanx ) ( 1 + sin2x ) = 1 + tanx
兩邊 × cosx ：
　　　≡　( cosx - sinx ) ( 1 + sin2x ) = cosx + sinx
兩邊 × ( cosx - sinx ) ：
　　　≡　( cosx - sinx )2 ( 1 + sin2x ) = cos2x - sin2x
　　　≡　( cos2x - 2 sinx cosx + sin2x ) ( 1 + sin2x ) = ( 1 - sin2x ) - sin2x
　　　≡　( 1 - sin2x) ( 1 + sin2x ) = 1 - 2 sin2x
　　　≡　1 - sin22x = 1 - 2 sin2x
　　　≡　2 sin2x = sin22x
　　　　　　　 　= 4 sin2x cos2x
　　　　 　 　 　1 = 2 cos2x
　　　　 　 cos2x = 1 / 2
　　　　　 　cosx = ± 1 / (√2 )
　　　　　 　　x = 135°，315°#　.........( 45°，225°不合 )
　 　 　　通解　 x = ( n - 1 / 4 ) π

2007-02-21 13:08:08 補充：
　　由於計算過程中，曾消去 (sinx)^2 項，但如此項等於零，則有 x = 0°，180°之解，因 0°≤ x ＜ 360°。通解為 x = ( n - 1 ) π

(1-tanx)(1+sin2x)=1+tanx
1 - sinx/cosx + 2sinxcosx - 2sin^2 x = 1+ sinx/cosx
2sinx cosx - 2sin^2 x - 2sinx/cosx = 0
sinx = 0.......................or ........................ 2cosx - 2sinx - 2/cosx = 0
x= 180 or 360...........................................cos^2 x - sinxcosx -1 = 0
...............................................................sinxcosx + 1 - cos^2 x=0
...............................................................sinxcos +sin^2 x = 0
............................................................sinx=0.............or............cosx+sinx=0
...........................................................(repeated)...................sin(90-x)+sinx=0
.........................................................................................2sin45cos(45-x)=0
..................................................................................................cos(x-45)=0
.............................................................................................x-45=90, 270
.................................................................................................x=135, 315
i.e. overall solution is x=135,180,315,360

2007-02-20 23:40:42 補充：
i am overlooking the question,360 is not the required answer,if you want, the general solutin isx=n(pi)............. or.................2n(pi) 3(pi)/4..........or .............2n(pi)-(pi)/4where n is any integers

(1-tanx)(1+sin2x)=1+tanx
1-sin2xtanx-tanx+sin2x-1-tanx=0
2sinxcosx-2sinxcosx(sinx/cosx)-sinxcosx=0
2sinxcos^2x-2sin^2x-sinxcos^2x=0
2sinx(1-sin^2x)-2sin^2x-sinx(1-sin^2x)=0
2sinx-2sin^3x-2sin^2x-sinx+sin^3x=0
sin^3x+2sin^2x-sinx=0
sinx(sin^2x+2sinx-1)=0
sinx=0 or sinx=[-2+/-(8)^0.5]/2
x=0,180,360 or x=24.47 , sinx=-2.414(rejected) (corr. to 4 sig. fig.)

2007-02-20 22:42:29 補充：
sorry, x=/=360

2007-02-21 00:43:05 補充：
2sinxcosx-2sinxcosx(sinx/cosx)-sinxcosx=0呢一步做錯左．．．．