`tan^2 (x) + tan(x) - 12 = 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 63 - Precalculus (3rd Edition, Ron Larson).
See all solutions for this textbook.

2 Answers | Add Yours

gsarora17's profile pic

gsarora17 | (Level 2) Associate Educator

Posted on

`tan^2(x)+tan(x)-12=0`

using quadratic formula,

`tan(x)=(-1+-sqrt(1^2-4*1*(-12)))/2`

`tan(x)=(-1+-sqrt(49))/2=(-1+-7)/2=3,-4`

solutions for tan(x)=3 for the range `0<=x<=2pi`

`x=arctan(3) , pi+arctan(3)`

solutions for tan(x)=-4 for the range `0<=x<=2pi`

`x=pi-arctan(4) , 2pi-arctan(4)`

Solutions are,

x `~~`  1.25 , 1.815 ,4.4 , 4.957 radians

Images:
This image has been Flagged as inappropriate Click to unflag
Image (1 of 1)
scisser's profile pic

scisser | (Level 3) Honors

Posted on

Factor

`( tan x + 4 ) ( tan x - 3 ) = 0 `

Set each part equal to 0


`tan x = - 4 , tan x = 3 `

Use arctan for solve for x


`x = 104 ^o , 284 ^o , 72^o , 252 ^o`

We’ve answered 318,916 questions. We can answer yours, too.

Ask a question