## Precalculus (6th Edition) Blitzer

The solutions to the given equation are $\frac{\pi }{4},\frac{3\pi }{4},\frac{5\pi }{4},\ \text{ and }\ \frac{7\pi }{4}$.
We have to solve the equation on the interval $[0,2\pi )$: \begin{align} & 5\text{se}{{\text{c}}^{2}}x-10=0 \\ & \text{se}{{\text{c}}^{2}}x=2 \end{align} By using the reciprocal identity of trigonometry ${{\sec }^{2}}x=\frac{1}{{{\cos }^{2}}x}$. \begin{align} & \frac{1}{{{\cos }^{2}}x}=2 \\ & \text{co}{{\text{s}}^{2}}x=\frac{1}{2} \\ & \text{cos}\ x=\sqrt{\frac{1}{2}} \end{align} Multiply 2 with the numerator and the denominator of the given expression: \begin{align} & \text{cos}\ x=\sqrt{\frac{1}{2}\times \frac{2}{2}} \\ & =\sqrt{\frac{2}{4}} \\ & =\frac{\sqrt{2}}{2} \end{align} After that, simplify it by solving the equation of $\frac{\sqrt{2}}{2}$. And take the expression both as a negative and positive value. $\text{cos}\ x=\pm \frac{\sqrt{2}}{2}$ Each factor needs to be calculated as: $\text{cos}\ x=\frac{\sqrt{2}}{2}$ Or $\text{cos}\ x=-\frac{\sqrt{2}}{2}$ So, in the quadrant graph, the value of cosine is $\frac{\sqrt{2}}{2}$ at $\frac{\pi }{4}$ , and $\frac{7\pi }{4}$. This implies, \begin{align} & \text{cos}\ x=\cos \frac{\pi }{4} \\ & x=\frac{\pi }{4} \end{align} Then, take the value of $x$ as $\frac{7\pi }{4}$: \begin{align} & \text{cos}\ x=\cos \frac{7\pi }{4} \\ & x=\frac{7\pi }{4} \end{align} And, So, in the quadrant graph, the value of cosine is $-\frac{\sqrt{2}}{2}$ at $\frac{3\pi }{4}$ , and $\frac{5\pi }{4}$. This implies, \begin{align} & \text{cos}\ x=\cos \frac{3\pi }{4} \\ & x=\frac{3\pi }{4} \end{align} Now, take the value of $x$ as $\frac{5\pi }{4}$: \begin{align} & \text{cos}\ x=\cos \frac{5\pi }{4} \\ & x=\frac{5\pi }{4} \end{align} These are the proposed solutions of the tangent functions. Thus, the actual solution in the interval $[0,2\pi )$ will be $\frac{\pi }{4},\frac{3\pi }{4},\frac{5\pi }{4},\ \text{ and }\ \frac{7\pi }{4}$.