Answer
The solutions to the given equation are $\frac{\pi }{6},\frac{\pi }{2},\frac{5\pi }{6},\text{ and }\ \frac{3\pi }{2}$.
Work Step by Step
We have to solve the equation on the interval $[0,2\pi )$:
$\begin{align}
& \text{cos}\ x\ \text{csc}\ x=2\text{cos}\ x \\
& \text{cos}\ x\ \text{csc}\ x-2\text{cos}\ x=0 \\
& cos\ x\left( \text{csc}\ x-2 \right)=0
\end{align}$
And each factor needs to be calculated as:
$\text{cos}\ x=0\ $
or
$\begin{align}
& \text{csc}\ x-2=0 \\
& \csc \ x=0+2 \\
& =2
\end{align}$
Then, solve for $x$ on the interval $[0,2\pi )$.
So, in the quadrant graph, the value of cosine is $0$ at $\frac{\pi }{2}$ and $\frac{3\pi }{2}$. This implies,
$\begin{align}
& \text{cos}x=\cos \frac{\pi }{2} \\
& x=\ \frac{\pi }{2}
\end{align}$
$\begin{align}
& \text{cos}x=\cos \frac{3\pi }{2} \\
& x=\ \frac{3\pi }{2}
\end{align}$
And,
The value of cosecant is not shown in the quadrant graph. By using the reciprocal identity of trigonometry, $\csc x=\frac{1}{\sin x}$, the cosecant function gets converted into the sine function as,
$\begin{align}
& \frac{1}{\sin x}=2 \\
& \sin x=\frac{1}{2}
\end{align}$
So, in the quadrant graph, the value of sine is $\frac{1}{2}$ at $\frac{\pi }{6}$ and $\frac{5\pi }{6}$. This implies,
$\begin{align}
& \sin x=\sin \frac{\pi }{6} \\
& x=\frac{\pi }{6}
\end{align}$
$\begin{align}
& \sin x=\sin \frac{5\pi }{6} \\
& x=\frac{5\pi }{6}
\end{align}$
These are the proposed solutions of the cosine and sine functions. Thus, the actual solutions in the interval $[0,2\pi )$ will be $\frac{\pi }{6},\frac{\pi }{2},\frac{5\pi }{6},\text{ and }\ \frac{3\pi }{2}$.
