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

Question

solve for x in radians, between 0 and 2pi

Answer 1

let a be cos2x | a | <= 1
4a^2 - 4a + 1 = 0
<=> (2a - 1)^2 = 0
<=> 2a -1 = 0
<=> a=1/2
<=> cos2x = cos (pi/3)
<=> 2x = +- pi/3 + k2pi (k is a subset of Z )
<=> x = +- pi/6 + kpi
according to the problem, we have these following roots :
x = pi/6 or x = 7pi/6 or x = -pi/6 or x = 5pi/6

Answer 2

4cos^2[2x]-4cos[2x]+1=0
From inspection we find the
expression is a perfect square.
{2cos[2x]-1}^2=0
2cos2x-1=0
2cos2x=1
cos2x=1/2
2x=Pi/3 rad.
x=Pi/6 or 30Deg

Answer 3

first, let cos(2x) = 't'
then the given equation can be written as :-
= 4t^2-4t+1=0
= 4t^2-2t-2t+1=0
= (2t-1)^2=0
= 2t-1 =0
= t=1/2
= cos(x)=1/2
therefore, x=60 degrees.

Answer 4

4cos^2(2x)-4cos(2x)+1=0
=>[2cos(2x) - 1]^2 = 0
=>2cos(2x) - 1=0
=>cos(2x) =1/2
=>cos(2x) =cos(pi/3)

writing general solution,

2x= 2n(pi) (+-) pi/3
=>x=n(pi) (+-) pi/6

put n=0,1,2...

pi/6, 5pi/6, 7pi/6, 11pi/6