Question

Precalculus only right answer please and thank you. Help me

Answer 1

Given:

$\frac{2tanx}{1-tan^2(x)} =\sqrt{3}$

We know the identity

$tan(2x)= \frac{2tanx}{1-tan^2(x)}$

So we can equate $tan(2x) =\sqrt{3}$

$tan(x) =\sqrt{3}$ when $x=\frac{\pi}{3}$

Tan is positive in first and third quadrant

So we will get one move value for x

$tan(x) =\sqrt{3}$ when $x=\frac{4\pi}{3}$

So for  $tan(2x) =\sqrt{3}$

$2x=\frac{4\pi}{3}$  and $2x=\frac{\pi}{3}$

Divide by 2 on both sides

$x=\frac{2\pi}{3}$  and $2x=\frac{\pi}{6}$

To get general solution we add $n\pi$

So option A  and option C are correct.