# `tan^2 (x) + tan(x) - 12 = 0` Use inverse functions where needed to find all solutions of the equation in the interval `0,2pi)`.

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Expert Answers

gsarora17 | Certified Educator

`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**

Student Comments

scisser | Student

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`