## Precalculus (6th Edition) Blitzer

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