A Maths問題

1.cos x +cos 3x=0
2cos2xcox=0
cos2x=0 or cosx=0
2x=90˚,270˚,450˚,630˚ or x=90˚,270˚
x=45˚,90˚,135˚,225˚,270˚ or 315˚

2.sin3x+cos3x=sin2x+cos2x
sin3x-sin2x=cos2x-cos3x
2cos(5x/2)sin(x/2)=-2sin(5x/2)sin(-x/2)
cos(5x/2)sin(x/2)=sin(5x/2)sin(x/2)
cos(5x/2)sin(x/2)-sin(5x/2)sin(x/2)=0
sin(x/2)[cos(5x/2)-sin(5x/2)]=0
sin(x/2)=0 or tan(5x/2)=1
x/2=0˚,180˚ or 5x/2=45˚,225˚,405˚,585˚,765˚
x=0˚,18˚,90˚,162˚,234˚,306˚ or 360˚

3.cos (x+15˚)cos(x-15˚)-sinx sin(x-30˚)=cos165˚
1/2[cos2x+cos30˚]+1/2[cos(2x-30˚)-cos30˚]=cos165˚
cos2x+cos30˚+cos(2x-30˚)-cos30˚=2cos165˚
cos2x+cos(2x-30˚)=2cos165˚
2cos(2x-15˚)cos15˚=2cos(180˚-15˚)
cos(2x-15˚)cos15˚=-cos15˚
cos(2x-15˚)=-1
2x-15˚=270˚ or 630˚
2x=285˚ or 645˚
x=142.5˚ or 322.5˚

4.cos x+cos2x+cos3x+cos4x=0
cosx+cos3x+cos2x+cos4x=0
2cos2xcosx+2cos3xcosx=0
cosx(cos2x+cos3x)=0
cosx[2cos(5x/2)cos(x/2)]=0
cosx=0 or cos(5x/2)=0 or cos(x/2)=0
x=90˚,270˚ or 5x/2=90˚,270˚,450˚,630˚,810˚ or x/2=90˚
x=36˚,90˚,180˚,252˚,270˚,324˚

5. tan(30˚+x) tan(30˚-x) = 1 - 2cos2x
sin(30˚+x)sin(30˚-x)/cos(30˚+x)cos(30˚-x) = 1 - 2cos2x
(1/2)(cos2x-cos60˚)/(1/2)(cos60*+cos2x) = 1 - 2cos2x
cos2x- 1/2 = (1-2cos2x)( 1/2 + cos2x )
cos2x-1/2 = 1/2+cos2x-cos2x-2cos22x
2cos22x+cos2x-1=0
(2cos2x-1)(cos2x+1)=0
cos2x=1/2 or cos2x=-1
2x = 60˚, 300˚,420˚, 660˚ or 2x = 180˚, 540˚
x = 30˚, 150˚, 210˚, 330˚ or 90˚, 270˚

Q.5
tan(30 +x)tan(30-x)= sin(30+x)sin(30-x)/cos(30+x)cos(30-x)=(cos2x-cos60)/2 divided by (cos60 + cos2x)/2 = (cos2x-1/2)/(1/2 +cos2x) = (2cos2x - 1)/(1+2cos2x), which is equal to 1-2cos2x according to the equation.
That is (2cos2x - 1) = (1-2cos2x)(1 + 2cos2x)
(2cos2x - 1) -(1-2cos2x)(1+2cos2x) = 0
(2cos2x - 1) + (2cos2x - 1)(1 + 2cos2x) = 0
(2cos2x - 1)(1 + 1 + 2cos2x) = 0
Therefore, cos2x = 1/2...........(1) and
cos2x = -1.............(2)
For (1) x=30, 150.
For (2) x=90.
Therefore, x = 30, 90 and 150.

2008-05-25 13:53:02 補充：
x can also = 210,270 and 330.