What are the values of X

Question

2tanx/1-tan^2x = 1/√3

I know it is pi/6 + pin/2 but the answers to choose are:

7pi/12 +npi,

11pi/12 +npi,

5pi/12 +npi and

pi/12 +npi.

Are the correct answers 7pi/12 +npi and pi/12 +npi

Thank you

Answer 1

I suspect the given answer was computed by rewriting 2tanx/(1-tan²x) = 1/√3 as a quadratic equation; cross-multiply and rearrange to get

tan²x + 2√3 tanx - 1 = 0

Then letting y = tan x gives

y² + 2√3 y - 1 = 0

Using the Quadratic Formula, the solutions are y = -2 - √3 (the inverse tangent of which is -5pi/12, which gives the 7pi/12 + n*pi solution set) and y = 2 - √3 (the inverse tangent of which is pi/12, which gives the other solutions).

Answer 2

The initial equation is:

(2 tan x)/(1 - tan^2 x) = 1/sqrt(3)

(sqrt is the square root)

Among your trigonometric identities, you should have the one that gives you the equivalent of doubling the angle of a tangent:

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

So you can use it to infer that:

tan (2x) = 1 / sqrt (3)

The angle that has the tangent 1 / sqrt(3) is pi/6

tan (pi/6) = 1 / sqrt(3)

You can thus infer that:

tan (2x) = tan (pi/6)

Therefore:

2x = pi /6 + k*pi

We add the k*pi because the function is periodic, and its period is pi.

x = pi/12 + k*pi/2

If we consider only the cases where k is even, then we can posit k = 2n and the equation becomes:

pi/12 + 2n*pi/2 = pi/12 + n*pi

If we consider only the cases where k is odd, then we can posit k = 2n+1 and the equation becomes:

pi/12 + (2n+1)*pi/2 = pi/12 + 2n*pi/2 + pi/2 = pi/12 + 6pi/12 + n*pi = 7pi/12 + n*pi

Answer 3

Yes, you got it right

'x= pi/12+npi and 7pi/12 +np, n=any integer

tanx =2-sqr3,-2-sqr3

x = arctan(2-sqr3), arctan(-2-sqr3) +npi

or

x= 15 degrees or 105 degrees plus 180n

Answer 4

2x = π/6 + nπ/2, n ∈ Z⁺

x = π/12 + 6nπ/12, n ∈ Z⁺

x = 7nπ/12, n ∈ Z⁺