# Find All Real Solutions (problem)

Problem is:

2 cos 3x - 1 = 0

Here's how I've done it so far:

2 cos 3x = 1

cos 3x = 1/2

Now I'm stuck. Where do I go from here? Do I divide the 3 out making it cos x = 1/6? I know how to find all the real solutions once I find what cos x is. I'm just having trouble figure out exactly how to get cos x in that type of problem.

Any help is grealy appreciated.

Last edited:

Originally posted by Cod
Do I divide the 3 out making it cos x = 3/2?
I think you meant cos x = 1/6. No, we can't take the 3 out.

I know how to find all the real solutions once I find what cos x is. I'm just having trouble figure out exactly how to get cos x in that type of problem.
let y = 3x
cos y = 1/2
y = [the general solution]
3x = [the general solution]
x = 1/3 * [the general solution]

just let me make the [general solution] a little bit clearer.
Now, we first define y=3x
therefore :
cos y = 1/2
or :
y = cos-1(1/2) + 2n[pi]
Now, you might be thinking "Where did 2n[pi] come from ?"
Well, you must remember that Cos() repeats itself after 2[pi], 4[pi], 6[pi], ... , or in other words after 2n[pi] (where n is an integer), so the value of cos(z) and cos(z+2n[pi]) is the same.
Now, from the last equation, you must either use a calculator to figure out cos-1(1/2), or try to remember if the number looks familiar.
The number looks familiar to me, i know that Cos([pi]/3) = 1/2 , or
Cos-1(1/2) = [pi]/3
So, now, you can rewrite your equation, and solve it.

I think it would be quite useful if you can draw a y = cos x graph.

Anyway, you just need to find all the solutions where cos x = 1/2. Then divide x by a factor of 3.