# Determine all the values of x if 2tan^2x-tanx-1=0

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To solve 2tan^2x-tanx -1 = 0.

To solve the problem we put tanx = t in the given equation:

2t^2-t -1 = 0

We factor the left by splitting the middle term -t as -2t +t:

2t^2-2t+t -1= 0

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

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

t- 1 = 0. Or 2t+1 =0

t =1 . Or t = -1/2.

Therefore t = 1 gives tanx = 1.

So x = pi/4 or 5pi/4. Or

t = -1/2 gives, tanx = -1/2.

x = -arc tan (-0.50) = -26.57 deg or 153.43 deg .

We have to solve 2(tan x)^x - tan x - 1 = 0

Now 2 (tan x)^2 - tan x - 1 = 0

=> 2 (tan x)^2 - 2tan x + tan x - 1 = 0

=> 2 tan x( tan x -1) +1 (tan x -1) =0

=> (2 tan x +1) (tan x-1) =0

Therefore tan x = -1/2 or tan x =1

If tan x = 1 , x = 45 degree

If tan x = -1/2 , x = -26.56 degree

**Therefore the solutions are 45 degree and -26.56 degree.**

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}**