Stuck on this hard trig problem

1. Aug 21, 2004

Short term memory

ok heres the problem i haven't taken trig for like 2 years and suddenly i have been thrown back in to it...

problem: 2 cos 2x - 4 cos x + 3 = 0 0 < x < 2 pi

my work (used cos 2 x = 2 cos^2 x - 1)

4 cos^2 x - 2 - 4 cos x + 3 = 0

4 cos^2 x - 4 cos x + 1 = 0

ok now here is my problem if i try and factor it out i ll be left with the a really ugly 1/ ( 4 cos x)

is there another equation i could use... is my logic flawed... is there an easier way... do i just have to do it the ugly way... ???

2. Aug 21, 2004

3.14159265358979

2cos2x - 4cosx + 3 = 0
2(2cos^{2}x-1) - 4cosx + 3 = 0
4cos^{2}x - 4cosx + 1 = 0
(2cosx-1)^{2} = 0
cosx = 1\2
x = 60 deg. or 300 deg.

3. Aug 21, 2004

Short term memory

... me stupid...

PS having a little to much fun with pi arn't we... you should go out to atleast when pi gets to 123456789......

4. Aug 22, 2004

HallsofIvy

Staff Emeritus

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