Solve for x
[tex]\frac{2\tan (x)}{1-tan^2(x)}=\frac{1}{\sqrt[]{3}}[/tex]
Step 1: Let tan(x) represent y
[tex]\begin{gathered} \frac{2y}{1-y^2}=\frac{1}{\sqrt[]{3}} \\ \text{cross multiply} \\ 1-y^2=2\sqrt[]{3}y \\ y^2+2\sqrt[]{3}y-1=0 \end{gathered}[/tex]
Step 2: Using quadratic equation solve for y
[tex]\begin{gathered} y^2+2\sqrt[]{3}y-1=0 \\ y=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ a=1,b=2\sqrt[]{3},c=-1 \\ y=\frac{-2\sqrt[]{3}\pm\sqrt[]{(2\sqrt[]{3})^2-4(1)(-1)}}{2(1)} \\ y=\frac{-2\sqrt[]{3}\pm\sqrt[]{12+4}}{2} \\ y=\frac{-2\sqrt[]{3}\pm\sqrt[]{16}}{2} \\ y=\frac{2(-\sqrt[]{3}\pm2)}{2} \\ y=-\sqrt[]{3}\pm2 \end{gathered}[/tex]
Since the y represents tan x
[tex]\begin{gathered} \tan \mleft(x\mright)=-\sqrt{3}-2,\: \tan \mleft(x\mright)=2-\sqrt{3} \\ x=\arctan \mleft(-\sqrt{3}-2\mright)+\pi n,\: x=\arctan \mleft(2-\sqrt{3}\mright)+\pi n \\ x\text{ in degrees} \\ \: x=-75^{\circ\: }+180^{\circ\: }n,\: x=15^{\circ\: }+180^{\circ\: }n \\ \: x=15^{\circ\: }+180^{\circ\: }n \end{gathered}[/tex]
Therefore the correct answer is
[tex]x=\frac{\pi}{12}+n\pi[/tex]
Hence the correct answer is Option B
