3^1/2 tan x *(3^1/2*tanx - 1)= 0
tan x = 0 or 1/sqrt3
x = 0, pi, pi/6 and one other tan^-1(1/sqrt3)
Solve for all real values of x.
3tan^2x = 3^1/2 tan x
I have no idea how to approach this problem.
3 answers
3tan^2x = 3^1/2 tan x
3tan^2x - √3tan x = 0
tanx(3tanx - √3) = 0
so tanx = 0 or tanx = √3/3
if tanx = 0, then x = 0º, 180º, 360º, ...
if tanx = √3/3 then x = 30º, 210º,390º, ...
The period of the tangent function is 180º, so adding 180 to any of our answers produces more answers
the angles in radians would be
0, pi, 2pi ...
pi/6, 7pi/6, 13pi/6, ..
(adding pi each time)
3tan^2x - √3tan x = 0
tanx(3tanx - √3) = 0
so tanx = 0 or tanx = √3/3
if tanx = 0, then x = 0º, 180º, 360º, ...
if tanx = √3/3 then x = 30º, 210º,390º, ...
The period of the tangent function is 180º, so adding 180 to any of our answers produces more answers
the angles in radians would be
0, pi, 2pi ...
pi/6, 7pi/6, 13pi/6, ..
(adding pi each time)
Much appreciated guys.
3 answers