Solving Trigonometric Equation: Alternate Methods?

[Short term memory]

ok here's the problem i haven't taken trig for like 2 years and suddenly i have been thrown back into 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... ?

 

[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.

 

[Short term memory]

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.

... 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...

 

[HallsofIvy]

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

"There is not enough room in this margin..."