Question

If 2 tan x 1- tan2 x 1 V3 then x can equal: + MT B. X= A. x=112 x- 7 C. X = ST D. X=

Answer 1

Solve for x

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

Step 1: Let tan(x) represent y

$\begin{gathered} \frac{2y}{1-y^2}=\frac{1}{\sqrt[]{3}} \\ \text{cross multiply} \\ 1-y^2=2\sqrt[]{3}y \\ y^2+2\sqrt[]{3}y-1=0 \end{gathered}$

Step 2: Using quadratic equation solve for y

$\begin{gathered} y^2+2\sqrt[]{3}y-1=0 \\ y=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ a=1,b=2\sqrt[]{3},c=-1 \\ y=\frac{-2\sqrt[]{3}\pm\sqrt[]{(2\sqrt[]{3})^2-4(1)(-1)}}{2(1)} \\ y=\frac{-2\sqrt[]{3}\pm\sqrt[]{12+4}}{2} \\ y=\frac{-2\sqrt[]{3}\pm\sqrt[]{16}}{2} \\ y=\frac{2(-\sqrt[]{3}\pm2)}{2} \\ y=-\sqrt[]{3}\pm2 \end{gathered}$

Since the y represents tan x

$\begin{gathered} \tan \mleft(x\mright)=-\sqrt{3}-2,\: \tan \mleft(x\mright)=2-\sqrt{3} \\ x=\arctan \mleft(-\sqrt{3}-2\mright)+\pi n,\: x=\arctan \mleft(2-\sqrt{3}\mright)+\pi n \\ x\text{ in degrees} \\ \: x=-75^{\circ\: }+180^{\circ\: }n,\: x=15^{\circ\: }+180^{\circ\: }n \\ \: x=15^{\circ\: }+180^{\circ\: }n \end{gathered}$

Therefore the correct answer is

$x=\frac{\pi}{12}+n\pi$

Hence the correct answer is Option B