# How do you solve 2cos^3x + cos^2x = 0 in the interval [0, 2pi]?

##### 1 Answer
Apr 25, 2016

$x \in \left\{\frac{\pi}{2} , \frac{2 \pi}{3} , \frac{4 \pi}{3} , \frac{3 \pi}{2}\right\}$

#### Explanation:

$2 {\cos}^{3} x + {\cos}^{2} x = 0$
$2 {\cos}^{2} x \left(\cos x + \frac{1}{2}\right) = 0$
${\cos}^{2} x = 0 \mathmr{and} \cos x + \frac{1}{2} = 0$
$\cos x = 0 \mathmr{and} \cos x = - \frac{1}{2}$
$x = \frac{\pi}{2} + k \pi \mathmr{and} \left(x = \frac{2 \pi}{3} + 2 k \pi \mathmr{and} x = \frac{4 \pi}{3} + 2 k \pi\right)$, where $k \in m a t h \boldsymbol{Z}$
Those are all real solutions but since we care about only the interval $\left[0 , 2 \pi\right]$ the final solutions are: $\frac{\pi}{2} , \frac{2 \pi}{3} , \frac{4 \pi}{3} , \frac{3 \pi}{2}$.