## Algebra and Trigonometry 10th Edition

$\alpha = \dfrac{\pi}{6},\dfrac{\pi}{2}, \dfrac{5\pi}{6},\dfrac{3\pi}{2}$
$2 cos \alpha \sin \alpha -\cos \alpha= 0$ 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: $\alpha = \dfrac{\pi}{6},\dfrac{\pi}{2}, \dfrac{5\pi}{6},\dfrac{3\pi}{2}$