# `sec^2 (x) + tan(x) - 3 = 0` Use inverse functions where needed to find all solutions of the equation in the interval `0,2pi)`.

### Textbook Question

Chapter 5, 5.3 - Problem 66 - Precalculus (3rd Edition, Ron Larson).
See all solutions for this textbook.

gsarora17 | (Level 2) Associate Educator

Posted on

`sec^2(x)+tan(x)-3=0`

use the identity `sec^2(x)=1+tan^2(x)`

`1+tan^2(x)+tan(x)-3=0`

`tan^2(x)+tan(x)-2=0`

`tan(x)=(-1+-sqrt(1^2-4*1*(-2)))/2`

`tan(x)=(-1+-sqrt(9))/2=(-1+-3)/2=1,-2`

General solutions for tan(x) =1 are x=pi/4+pin

Solutions for the `0<=x<=2pi`  range are,

x=pi/4 , x=(5pi)/4

Solutions for tan(x)=-2 for the range  are,

`x=pi-arctan(2) , 2pi-arctan(2)`

Solutions are

x= `pi/4` , `(5pi)/4`, 2.034 , 5.176 radians

scisser | (Level 3) Honors

Posted on

`sec^2x + tanx - 3 = 0 `

Change `sec^2x` to `tan^2x+1`
`tan^2x + 1 + tanx - 3 = 0 `

Group like terms
`tan^2x + tanx - 2 = 0 `

Factor
`(tanx + 2)(tanx - 1) = 0 `

Set each part equal to 0
`tanx = -2 or tanx = 1 `

Use arctan to solve for x