Answer
$\alpha = \dfrac{\pi}{6},\dfrac{\pi}{2}, \dfrac{5\pi}{6},\dfrac{3\pi}{2}$
Work Step by Step
$2 cos \alpha \sin \alpha -\cos \alpha= 0$
We can simplify the equation as follows:
$ cos \alpha (2 \sin \alpha - 1)= 0$
$cos \alpha =0 $ and $\sin \alpha= \dfrac{1}{2}$
This gives 4 possible solutions:
$\alpha = \dfrac{\pi}{6},\dfrac{\pi}{2}, \dfrac{5\pi}{6},\dfrac{3\pi}{2}$
