Trig solve for x [ Attempt Included ]

[lovemake1]

Homework Statement

solve the following equation for x in the interval [0,2pi)
2Cos^2(2x) + 1 = 3Cos(2x)

Homework Equations

Cos(2x) = 2Cos^2 (x) - 1

The Attempt at a Solution

2Cos^2(2x) + 1 = 3Cos(2x)

2Cos(2x)Cos(2x) + 1 = 3Cos(2x)
2Cos(2x) + 1 = 3
2(2Cos^2 (x) - 1) - 2 =0
4Cos^2 (x) - 2 - 2 = 0
4cos^2 (x) = 4
Cos^2 (x) = 1
Cosx = 1

x = 0 is this right?
please correct me if i messed up thanks,

 

[Shaggy16]

That's one answer, but there are more. Try the substitution u=cos(2x). That will give you a quadratic, which will give you both of the answers you need.

 

[lovemake1]

2cos^2(2x) + 1 = 3 cos(2x)
2cos^2(2x) - 3cos(2x) + 1 = 0

let u = cos(2x)

2u^2 - 3u + 1 = 0
(2u - 1 )(u - 1) = 0

u = 1/2, 1

Case 1. Cos2x = 1/2
cos2x = 1/2
2cos^2x - 1 = 1/2
2cos^2x = 1/2 - 1
2cos^2x - = -1/2
cos^2x = -1/4
cosx = sqrt(-1/4)

but how can you sqrt negative? is this error ?


Case 2. Cos2x = 1
2Cos^2x - 1 = 1
2Cos^2x - 2 =0
2(Cos^2x - 1) = 0
Cos^2x = 1
Cosx = 1


Did i get it right ?

 

[Shaggy16]

Case 2 is correct, but in case 1, you should have added 1 instead of subtracting it.

Case 1:
cos(2x)=1/2
2cos^2(2x)-1=1/2
2cos^2(2x)=1/2+1=3/2
Try it from here.

And you have to remember that the function is periodic, so there will be multiple solutions to each of those.