How do I find the exact solutions? (0, 2pi)
3cos x -2= cos 2x

Question

How do I find the exact solutions? (0, 2pi)
               3cos x -2= cos 2x

tkhunny · Answer

I'm thinking $\cos(2x) = 2\cos^{2}(x) - 1$ might be useful.

anonymous · Answer

Then would I just set each side to zero?

tkhunny · Answer

You do not know why you set things to zero, do you?

The idea is this, if a*b = 0, then either a = 0 or b = 0 (or both.)  If you don't have multiplication, it doesn't do the same thing.  If you don't have zero, it doesn't do anything.  You need two things multiplied together to make zero.

$3\cos x - 2= \cos 2x$

$3\cos x - 2= 2\cos^{2}(x) - 1$

$2\cos^{2}(x) - 3\cos x + 1 = 0$

$(2\cos(x) - 1)(\cos(x) - 1) = 0$

NOW we can start setting things to zero!

anonymous · Answer

would the solutions end up being pi/3, 5pi/3, pi/3?