Given:
[tex]\frac{2tanx}{1-tan^2(x)} =\sqrt{3}[/tex]
We know the identity
[tex]tan(2x)= \frac{2tanx}{1-tan^2(x)}[/tex]
So we can equate [tex]tan(2x) =\sqrt{3}[/tex]
[tex]tan(x) =\sqrt{3}[/tex] when [tex]x=\frac{\pi}{3}[/tex]
Tan is positive in first and third quadrant
So we will get one move value for x
[tex]tan(x) =\sqrt{3}[/tex] when [tex]x=\frac{4\pi}{3}[/tex]
So for [tex]tan(2x) =\sqrt{3}[/tex]
[tex]2x=\frac{4\pi}{3}[/tex] and [tex]2x=\frac{\pi}{3}[/tex]
Divide by 2 on both sides
[tex]x=\frac{2\pi}{3}[/tex] and [tex]2x=\frac{\pi}{6}[/tex]
To get general solution we add [tex]n\pi[/tex]
So option A and option C are correct.
