Answer
The solutions in the interval $[0,2\pi )$ are $\frac{2\pi }{3},\,\ \pi,\,\ \text{ and }\ \text{ }\!\!~\!\!\text{ }\frac{4\pi }{3}$.
Work Step by Step
We have to solve the equation on the interval $[0,2\pi )$; the following will be the course of action. Start with:
$2{{\cos }^{2}}x+3\cos x+1=0$
Solve it as
$\begin{align}
& 2{{\cos }^{2}}x+2\cos x+\cos x+1=0 \\
& 2\cos x\left( \cos x+1 \right)+1\left( \cos x+1 \right)=0 \\
& \left( 2\cos x+1 \right)\left( \cos x+1 \right)=0
\end{align}$
Each factor needs to be calculated as follows:
$\begin{align}
& 2\cos x+1=0 \\
& 2\cos x=0-1 \\
& \cos x=-\frac{1}{2}
\end{align}$
or
$\begin{align}
& \cos x+1=0 \\
& \cos x=0-1 \\
& \cos x=-1
\end{align}$
Then, solve for $x$ on the interval $[0,2\pi )$.
In the quadrant graph, the value of cosine is $-\frac{1}{2}$ at the angle of $\frac{2\pi }{3}$ and $\frac{4\pi }{3}$.
It implies,
$\begin{align}
& \cos x=\cos \frac{2\pi }{3} \\
& x=\frac{2\pi }{3}
\end{align}$
$\begin{align}
& \cos x=\cos \frac{4\pi }{3} \\
& x=\frac{4\pi }{3}
\end{align}$
In the quadrant graph, the value of cosine is $-1$ at the angle of $\pi $. It implies,
$\begin{align}
& \cos x=\cos \pi \\
& x=\pi
\end{align}$
These are the proposed solutions of the cosine function. Thus, the actual solutions in the interval $[0,2\pi )$ will be $\frac{2\pi }{3},\ \,\pi,\ \,\text{ and }\ \text{ }\!\!~\!\!\text{ }\frac{4\pi }{3}$.
