# HELP Trig Equation

1. Jun 10, 2006

### 01suite

soo okay...i have this question that im stuck on...

Solve for x,

cos 2x - cos^2x = 0, -180degrees<x<180degrees

i tried solving it...

cos 2x - cos^2x = 0
[2cos^2x -1]-cos^2x = 0
cos^2x -1 = 0 <--(???)
(cos x -1)(cos x +1) = 0
therefore... 1)cos x = -1 , 2)cos x = 1

so...

1)cos x = -1
x = cos^-1 (-1)
x= 180

2) cos x = 1
x = 0

but i have tried on my calculator...and -180 degrees can be one of the answers too...i dont get how to get -180

2. Jun 10, 2006

### dav2008

Well where is cos(x) equal to -1 besides x=180°?

(Also, since the original problem asks for solutions greater than -180° and less than 180°, then neither 180° nor -180° would be solutions. Maybe the problem had -180° ≤ x ≤ 180°)

Last edited: Jun 10, 2006
3. Jun 10, 2006

### 01suite

the thing is...i dont even know if i did the equation right.... T.T it seems wrong

4. Jun 10, 2006

### dav2008

How does it "seem wrong"? Either you made a mistake and it's wrong or you didn't and it's right.

I didn't see any mistakes except for the two things I pointed out.

You can plug in the values of x you found into the original equation to check your work.

Last edited: Jun 10, 2006
5. Jun 10, 2006

### 01suite

umm..thanks..

6. Jun 11, 2006

### TD

You're doing all right, the solution are $k\pi$ with $k \in \mathbb{Z}$.
The only solution in the given interval is x = 0, if the inequalities weren't strict, you have to add pi and -pi as solutions.