SOLVE EQUATION FOR 0度 大過or等如 x 大過or等如 360度

19.
cos2x = sin^4 x + cos^4 x
1-2sin^2 x=sin^4 x + (1-sin^2 x)^2
1-2sin^2 x=sin^4 x+1-2sin^2 x+sin^4x
2sin^4 x=0
x=0or180

2007-03-21 18:13:17 補充：
cos2x=cos^2 x - sin^2 x = 2cos^2 x - 1 = 1 - 2sin^2 xsin^2 x + cos^2 x = 1

2007-03-21 18:18:58 補充：
19. cos2x = sin^4 x + cos^4 x2cos^2 x-1=(1-cos^2 x)^2+cos^4 x2cos^2 x-1=1-2cos^2 x+cos^4 x+cos^4x2cos^4 x-4cos^2 x+2=0cos^4 x-2cos^2 x+1=0(cos^2 x-1)^2=0(cos x+1)(cos x-1)=0cos x=-1or1x=0or180

2007-03-21 18:20:08 補充：
19.cos2x = sin^4 x cos^4 x1-2sin^2 x=sin^4 x (1-sin^2 x)^21-2sin^2 x=sin^4 x 1-2sin^2 x sin^4x2sin^4 x=0x=0or180or360

很简单

cos2x = sin^4 x + cos^4 x
cos2x = (1-cos^2x)^2 + cos^4x
cos2x = 1 - 2cos^2x + cos+ cos^4x
cos2x = 1 - 2cos^2x + 2cos^4x
2cos^2x-1 = 1 - 2cos^2x + 2cos^4x
4cos^2x - 2cos^4x - 2 = 0
-2(cos^4x - 2cos^2x + 1) = 0
cos^4x - 2cos^2x + 1 = 0
(cos^2x - 1)^2 = 0
cos^2x -1 = 0
cosx = 1 or -1

∴ x = 0 180 or 360

cos2x = sin^4x + cos^4x
2cos^2x-1 = (1-cos^2x)^2 + cos^4x
2cos^2x-1 = cos^4x - 2cos^2x + 1 + cos^4x
2cos^4x - 4cos^2x + 2 = 0
2(cos^4x - 2cos^2x + 1) = 0
cos^4x - 2cos^2x + 1 = 0
(cos^2x - 1)^2 = 0
cos^2x -1 = 0
cosx = 1 or -1

x = 0 or x = 180 or x = 360