## Precalculus (6th Edition) Blitzer

The solutions to the given equation are $\frac{\pi }{6},\,\frac{5\pi }{6},\,\frac{7\pi }{6},\ \text{ and }\ \frac{11\pi }{6}$.
We solve the equation on the interval $[0,2\pi )$: \begin{align} & 5\text{co}{{\text{t}}^{2}}x-15=0 \\ & \text{co}{{\text{t}}^{2}}x=3 \end{align} And using the reciprocal identity of trigonometry ${{\cot }^{2}}x=\frac{1}{{{\tan }^{2}}x}$, the above expression can be further simplified as: \begin{align} & \frac{1}{\text{ta}{{\text{n}}^{2}}x}=3 \\ & \text{ta}{{\text{n}}^{2}}x=\frac{1}{3} \\ & \text{tan}x=\pm \sqrt{\frac{1}{3}} \end{align} Simplifying it by solving the equation of $\frac{\sqrt{3}}{3}$ by taking the expression both as a negative and positive value. $\text{tan}\ x=\pm \frac{\sqrt{3}}{3}$ And each factor needs to be calculated as: $\text{tan}\ x=\frac{\sqrt{3}}{3}$ or $\text{tan}\ x=-\frac{\sqrt{3}}{3}$ So, in the quadrant graph, the value of tangent is $\frac{\sqrt{3}}{3}$ at $\frac{\pi }{6}$, and $\frac{7\pi }{6}$: \begin{align} & \text{tan}\ x=\tan \frac{\pi }{6} \\ & x=\frac{\pi }{6} \end{align} \begin{align} & \text{tan}\ x=\tan \frac{7\pi }{6} \\ & x=\frac{7\pi }{6} \end{align} And in the quadrant graph, the value of tangent is $-\frac{\sqrt{3}}{3}$ at $\frac{5\pi }{6}$ and $\frac{11\pi }{6}$: \begin{align} & \text{tan}\ x=\tan \frac{5\pi }{6} \\ & x=\frac{5\pi }{6} \end{align} \begin{align} & \text{tan}\ x=\tan \frac{11\pi }{6} \\ & x=\frac{11\pi }{6} \end{align} These are the proposed solutions of the tangent functions. Thus, the actual solutions in the interval $[0,2\pi )$ will be $\frac{\pi }{6},\,\frac{5\pi }{6},\,\frac{7\pi }{6},\ \text{ and }\ \frac{11\pi }{6}$.