opgave b

$
2\cos \left( {2x - \frac{1}{3}\pi } \right) = \sqrt 3
$
$
\cos \left( {2x - \frac{1}{3}\pi } \right) = \frac{1}{2}\sqrt 3
$
$
2x - \frac{1}{3}\pi  = \frac{1}{6}\pi  + k \cdot 2\pi
$ of $
2x - \frac{1}{3}\pi  =  - \frac{1}{6}\pi  + k \cdot 2\pi
$
$
2x = \frac{1}{2}\pi  + k \cdot 2\pi
$ of $
2x = \frac{1}{6}\pi  + k \cdot 2\pi
$
$
x = \frac{1}{4}\pi  + k \cdot \pi
$ of $
x = \frac{1}{{12}}\pi  + k \cdot \pi
$