For what value of x is the following 2(tan x)^2+tan(-x)=1 true

3 Answers | Add Yours

william1941's profile pic

william1941 | College Teacher | (Level 3) Valedictorian

Posted on

Now tan -x = - tan x

So 2(tan x)^2 + tan(-x)=1

=> 2(tan x)^2 - tan(x)=1

let Y = tan x

=> 2Y^2 – Y – 1 =0

=> 2Y^2 -2Y + Y -1 =0

=> 2Y(Y-1) +1(Y-1)=0

=> (2Y+1)(Y-1) = 0

Now if 2Y + 1 = 0

=> Y = -1/2

=> tan x = -1/2

If Y =1

=> tan x =1

Therefore tan x can be -1/2 and 1

x is arctan -1/2 = -26.56 degree

and arctan 1 = 45 degree

The required values of x are -26.56 degree and 45 degree.

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

For the beginning, we'll notice that one terms has the opposite variable, -x. Since the tangent function is odd, we'll write the term:

tan(-x) = - tan x

We'll re-write the given expression:

2 (tan x)^2 - tan x = 1

We'll factorize by tan x:

tan x(2 tan x - 1) = 1

We'll put tan x = 1

x = pi/4 + k*pi

2 tan x - 1 = 1

We'll add 1 both sides:

2 tan x = 2

We'll divide by 2:

tan x = 1

x = pi/4 + k*pi

or

tan x = -1

The tangent fucntion is negative when x is in the second or the fourth quadrant.

x = pi - pi/4

x = 3pi/4 + k*pi

x = 2pi - pi/4

x = 7pi/4 + k*pi

The solutions of the equation are:

{pi/4 + k*pi ; 3pi/4 + k*pi ; 7pi/4 + k*pi}

neela's profile pic

neela | High School Teacher | (Level 3) Valedictorian

Posted on

2(tanx)^2+tan(-x)=1

We can rewrite this as :

2t^2-t-1 = 0, where t = tanx

(2t+1)(t-1) = 0

2t+1 =0 or t -1 = 0

 2t = -1/2 . or t = 1

t = -1/2 or t = 1

Principal values:

 t = -1/2 => tanx = -1/2 , when = -26.57 deg or x = 153.43 de nearly.

t = 1 gives: tanx = 1. Or x = 45 deg or 225 deg

We’ve answered 318,915 questions. We can answer yours, too.

Ask a question