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

Question

02pie

Answer 1

(2cos x-1)(2cos x-1)
cos x=1/2
x=π/3 and 5π/3

Answer 2

One thing to note is that your interval should be written as

0 <= x < 2pi
(it means "0 is less than x, which is less than 2pi")
The way you have it written now mathematically makes no sense.

Anyway, back to your question, it's a quadratic in disguise, and if you don't believe me, let u = cosx.

Then we have

4u^2 - 4u + 1 = 0

This factors as normal.

(2u - 1)(2u - 1) = 0
Therefore, 2u - 1 = 0
Substitute u = cosx back, to get

2cosx - 1 = 0
2cosx = 1
cosx = 1/2

Where is cosx equal to 1/2? Mentally thinking back to the unit circle, it is equal to 1/2 at pi/3 and 5pi/3
So
x = pi/3 and 5pi/3

Answer 3

You should recognise the LHS as a perfect square: its (2 cos x - 1)^2.
So you need 2 cos x = 1, so cos x = 0.5.
That gives x = pi/3 or 5pi/3.

Answer 4

set M = cosx

then 4M^2 - 4M + 1 = 0
(2M -1)(2M - 1) = 0
M = 1/2

so cos x = 1/2

x = pi/3 or 5pi/3

Answer 5

4cos(x)^2 - 4cos(x) + 1 = 0
(2cos(x) - 1)(2cos(x) - 1) = 0
(2cos(x) - 1)^2 = 0
2cos(x) - 1 = 0
2cos(x) = 1
cos(x) = (1/2)
x = (pi/3) or (5pi/3)