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

[Ecc1213]

What is the the answer to 4cos^2(2x)-4cos2x+1=0 0<x<2pi

I started by useing the quadradic formula....

cos2x= -4+/-THE SQUARE ROOT OF (-4)^2-4(4)(1) ALL DIVIDED BY 2(4)

This gave me cos2x=-1/2

but I was told that it was 1/2

I was told that therefore...
2x=arccos(1/2)
x=arccos(1/2)/2

and that arccos(1/2)=pi/6 which is one of my answers but I don't understand how to get the other answers. 5pi/6, 7pi/6 and 11pi/6.
I was told to use a CAST diagram but don't understand how to interpret or write one that will help.

 

[Deleted member 4993]

What is the the answer to 4cos^2(2x)-4cos2x+1=0 0<x<2pi

I started by useing the quadradic formula....

cos2x= -4+/-THE SQUARE ROOT OF (-4)^2-4(4)(1) ALL DIVIDED BY 2(4)

The solution to a quadratic equation (Ax[sup:1fitogom]2[/sup:1fitogom] + Bx + C = 0)is:

x [sub:1fitogom]1,2[/sub:1fitogom] = [-B ± ?(B[sup:1fitogom]2[/sup:1fitogom] - 4AC)]/(2AC)

Since your B = -4

The answer should be +1/2

cos(2x) = 1/2

2x = ±?/3 + 2n? (n =0 , 1, 2, ....)

x = ±?/6 + n? = ?/6 , (?/6 + ?), (-?/6 + ?) and (-?/6 + 2?)

 

[Ecc1213]

Your said...
2x = ±?/3 + 2n? (n =0 , 1, 2, ....)

x = ±?/6 + n? = ?/6 , (?/6 + ?), (-?/6 + ?) and (-?/6 + 2?)

Ok. but what does,(?/6 + ?), (-?/6 + ?) and (-?/6 + 2?) Equal? do they equal the 5pi/6, 7pi/6 and 11pi/6?

 

[Deleted member 4993]

Your said...
2x = ±?/3 + 2n? (n =0 , 1, 2, ....)

x = ±?/6 + n? = ?/6 , (?/6 + ?), (-?/6 + ?) and (-?/6 + 2?)

Ok. but what does,(?/6 + ?), (-?/6 + ?) and (-?/6 + 2?) Equal? do they equal the 5pi/6, 7pi/6 and 11pi/6?

what does (x + x/6) equal to? or, (1 + 1/6) equal to?

what does (x - x/6) equal to? or, (1 - 1/6) equal to?

 

[Ecc1213]

Oh. Thanks.

so pi/6 = pi\6
pi\6 + pi= 7pi\6
pi\6-pi= 5pi\6
and pi\6 + 2pi = 11pi\6

these are the answers I was given. am I right?