# How do you solve  2cosx+1=0?

Jun 11, 2015

$S = \left\{x | x = \frac{2}{3} \pi + 2 k \pi \vee x = \frac{4}{3} \pi + 2 k \pi \text{ where } k \in \mathbb{Z}\right\}$

#### Explanation:

$2 \cos x + 1 = 0$
$\cos x = - \frac{1}{2}$
We remember the values for which $\cos x = - \frac{1}{2}$.
x=2/3pi=120° or x=4/3pi=240°.

But don't forget that cos(x) has a $2 \pi$ periodicity, so:
$S = \left\{x | x = \frac{2}{3} \pi + 2 k \pi \vee x = \frac{4}{3} \pi + 2 k \pi \text{ where } k \in \mathbb{Z}\right\}$