2006-09-14 01:24:02 · 6 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
whatever the x, cos values range from -1 to 1.
x can be any number from 0 to 360 or 0 to 2pi
cos3x=4cos(^3)x- 3cosx
cos2x=cos(^2)x-sin(^2)x=cos(^2)x-1+cos(^2)X=2cos(^2)x-1
4cos(^3)x+2cos(^2)x-2cosx=1/2 | :cosx
4cos(^2)+2cosx=2+1/2
4cos(^2)+2cosx=13/2 now cosx=T (T-variable)
4T^2+2T-13/2=0 rezolve as a normal ecuation, after that cosx=T where T is the value you find, and x=acosT
GOD I hate writing this stuff here.
All trigonometric formulas can be found here:
http://math.ournet.md/formule/trigonom/trigonom.html
2006-09-14 01:28:13 · answer #1 · answered by Λиδѓεy™ 6 · 0⤊ 0⤋
There are multiple solutions. Let y be cosx+cos2x+cos3x. Therefore y=-1/2. Plot y=-1/2 and cosx+cos2x+cos3x. Find the intersection between curve and line. One of the graphical answer should be about 0.8975979 in radians. So much for the no solution.
2006-09-14 01:47:37 · answer #2 · answered by Reny 2 · 0⤊ 0⤋
cos x + 4cos^3 x -3cos x + 2cos^2x -1 +1/2=0
cos x(4cos^2 x -2 + 2cos x) = 1/2
4cos x(2cos^2 x +cos x - 1) =1
this is all I have for now, maybe you need to consider the real value ranges, i.e
the first factor is -4<=4cos x<=4
and for the quadratic inside you can get a range by considering the discriminant, it would be advisable to show that these factors do not give unity.
imagine 1/(4cos x) =(2cos^2 x+cos x-1)
hope the lead helps, but I suppose complex solutions will exist.
2006-09-14 02:30:42 · answer #3 · answered by yasiru89 6 · 0⤊ 0⤋
cos x. 1-10+ 23 tec.= 2.550 cos.
The initial velocity, cosine 10-19 tec.
is the amount of space travelled... ok?
2006-09-14 02:22:13 · answer #4 · answered by happy_face95 1 · 0⤊ 0⤋
What about x=154.3°
Solutions 2pi/7, 4pi/7,6pi/7.......
2006-09-14 01:31:40 · answer #5 · answered by deflagrated 4 · 0⤊ 0⤋
Because it's jibberish ?
2006-09-14 01:27:32 · answer #6 · answered by Vinegar Taster 7 · 0⤊ 0⤋