2 cot^2 x - 6 = 0?

Question

Identify the x-values that are solutions of the equation.

Answer 1

cot x = 1 /tan x

So:

2/tan^2(x) - 6 = 0

Divide by 2, and you get 1/tan^2 x - 3 = 0

1/ tan^2 x= 3

1/3 = tan^2 x

tan x= +- sqrt (1/3)

Since cos pi/6 = sqrt 3/2 and sin pi/6 = 1/2, you will notice that tan (pi/6) = sin (pi/6)/cos(pi/6) = (1/2) / (sqrt 3/2)

Multiply by the inverse and you get that tan (pi/6) = (1/2) * (2/sqrt3) = 1/sqrt3

So, the solution for tan x = 1/sqrt 3 are x= pi/6 +k*pi

And the solution for tan x = -1/sqrt 3 are x= 5pi/6+k*pi (k is any possible integer number).

Ana

Answer 2

1/tan^2(x)=3 so tan^2 x= 1/3 and tan x= +- 1/3sqrt3
x= pi/6 +k*pi
x= 5pi/6+k*pi (k any integer)