## Precalculus (6th Edition) Blitzer

The solutions to the given equation are $\frac{\pi }{6},\frac{\pi }{2},\frac{5\pi }{6},\text{ and }\ \frac{3\pi }{2}$.
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 $\frac{\pi }{6},\frac{\pi }{2},\frac{5\pi }{6},\text{ and }\ \frac{3\pi }{2}$.