This time with a harder one. All the hints are really appreciated. Been banging my head to wall few days with this one...
So, here it is:
$\displaystyle 2\cos^2 2x = \cos 2x + 1 $
Thanks!
--
When you think you know it, you don't!
This time with a harder one. All the hints are really appreciated. Been banging my head to wall few days with this one...
So, here it is:
$\displaystyle 2\cos^2 2x = \cos 2x + 1 $
Thanks!
--
When you think you know it, you don't!
Note that you can rewrite it as $\displaystyle 2\cos^22x-\cos 2x-1=0$. Also observe that if you make a substitution, say $\displaystyle u=\cos 2x$, the equation becomes quadratic in form: $\displaystyle 2u^2-u-1=0$.
Solve this quadratic equation for u, and then backsubstitute $\displaystyle u=\cos 2x$ to find x.
Can you take it from here?