# How do you find the general solutions for cos(2x) + 5cos(x) + 3 = 0?

Aug 18, 2015

For $x \in \left[0 , 2 \pi\right]$
$\textcolor{w h i t e}{\text{XXXX}}$$x = \pi$

#### Explanation:

$\cos \left(2 x\right) = 2 {\cos}^{2} \left(x\right) - 1$$\textcolor{w h i t e}{\text{XXXX}}$(one of the double angle formulae)

$\cos \left(2 x\right) + 5 \cos \left(x\right) + 3 = 0$

$\Rightarrow$$\textcolor{w h i t e}{\text{XXXX}}$$2 {\cos}^{2} \left(x\right) + 5 \cos \left(x\right) + 3 = 0$

Factoring:
$\Rightarrow$$\textcolor{w h i t e}{\text{XXXX}}$$\left(2 \cos \left(x\right) + 3\right) \left(\cos \left(x\right) + 1\right) = 0$

So
$\cos \left(x\right) = - \frac{3}{2}$$\textcolor{w h i t e}{\text{XXXX}}$or$\textcolor{w h i t e}{\text{XXXX}}$$\cos \left(x\right) = - 1$

Since $\cos \left(x\right) \in \left[- 1 , + 1\right]$ for all values of $x$
$\textcolor{w h i t e}{\text{XXXX}}$$\cos \left(x\right) = - \frac{3}{2}$ is an extraneous result.

$\cos \left(x\right) = - 1$
$\Rightarrow$$\textcolor{w h i t e}{\text{XXXX}}$$x = 3 \pi$ within the range $\left[0 , 2 \pi\right]$

if the range is not restricted:
$\textcolor{w h i t e}{\text{XXXX}}$$x = \pi + 2 n \pi , \forall n \in \mathbb{Z}$

Note:
I modified a term in the question from ($\cos 2 \left(x\right)$) to ($\cos \left(2 x\right)$);
the other possible interpretation might have been (${\cos}^{2} \left(x\right)$)
but that version has no solutions for $x$