## Algebra and Trigonometry 10th Edition

$x=\frac{\pi}{2}$ $x=\frac{3\pi}{2}$
$2~cos^2x+3~cos~x=0$ $cos~x(2~cos~x+3)=0$ $cos~x=0$ $x=\frac{\pi}{2}$ or $x=\frac{3\pi}{2}$ $2~cos~x+3=0$ $cos~x=\frac{3}{2}$ Remember that $cos~x$ is a periodic function and that $-1\lt cos~x\lt1$. So, there is no solution for $cos~x=\frac{3}{2}\gt1$