Question

Find all solutions to the equation in the interval [0, 2π). (3 points)

cos 2x - cos x = 0

Answer 1

Use the identity \(\large \cos(2x) = 2\cos^2 x - 1\) to rewrite the equation as \(\large 2\cos^2x-\cos x - 1 = 0\). Now note that this is quadratic in \(\large \cos x\); i.e. if we were to make the substitution \(\large t=\cos x\), then the equation turns into the quadratic equation \(\large 2t^2-t-1 = 0\). At this point, solve the quadratic equation for t, the substitute back in \(\large \cos x\) and then finally solve for x.

Do you think you can take things from here? I hope this made sense! :-)

Answer 2

Maybe, I'll work it out and see what happens.

Answer 3

So, so far I have
cos2x-cosx=0
cos2x-2cos^2x-1=0
2t^2-t-1=0
t^2=1
t=sqrt1
cosx=sqrt1
is that right so far?

Answer 4

cos2x-cosx=0
cos2x=cosx if these is true, then
2x=x must be true, go from there....

Answer 5

Ok... But I don't really understand where to go or how I am supposed to get three points out of the on equation...

Answer 6

@J-Girl: You messed up on your algebra... :-/
\(2t^2-t-1 = 0 \implies (2t+1)(t-1) = 0\implies t=1\text{ or } t=-\dfrac{1}{2}\). So either \(\cos x = 1\implies x=0\) or \(\cos x= -\dfrac{1}{2}\implies x=\dfrac{2\pi}{3},\,\dfrac{4\pi}{3}\).

Does this make sense? :-)

Answer 7

@Christopher Toni I don't really understand what happens after t=1or t= -1/2 I don't really understand the the cosx=1 -> x=0...