# Chapter 7 - 7.3 - Solving Trigonometric Equations - 7.3 Exercises - Page 531: 62

$x=\frac{\pi}{4}+n\pi$ and $x=arctan(-2)+n\pi$, where $n$ is an integer.

#### Work Step by Step

$sec^2x=1+tan^2x$ $sec^2x+tan~x=3$ $1+tan^2x+tan~x-3=0$ $tan^2x+tan~x-2=0~~$ ($tan~x=2~tan~x-tan~x$) $tan^2x+2~tan~x-tan~x-2=0$ $tan~x(tan~x+2)-1(tan~x+2)=0$ $(tan~x-1)(tan~x+2)=0$ $tan~x-1=0$ $tan~x=1$ $x=arctan~1=\frac{\pi}{4}$ $tan~x+2=0$ $tan~x=-2$ $x=arctan(-2)$ The period of $tan~x$ is $\pi$. So, add multiples of $\pi$ to each solution to find the general solution: $x=\frac{\pi}{4}+n\pi$ and $x=arctan(-2)+n\pi$, where $n$ is an integer.

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