Challenge Question Help!!!

cos theta can be 1/2 and 1.

Why is it a 'challange questions'?

$\cos(2\theta) = 2\cos(\theta)\\ cos^2\theta-\sin^2\theta = 2\cos(\theta)\\ cos^2\theta-(1-\cos^2\theta) = 2\cos(\theta)\\ 2cos^2\theta-1 = 2\cos\theta\\ 2cos^2\theta-2\cos\theta-1 = 0\\$

Now let x=cos (theta)  and solve for x.

etc