Question

If $x+\frac{1}{x}=\sqrt{3}$, then $x$ equal:

• A. $\cos \frac{\pi }{3}+i\sin \frac{\pi }{3}$
• B. $\cos \frac{\pi }{2}+i\sin \frac{\pi }{2}$
• C. $\sin \frac{\pi }{6}+i\cos \frac{\pi }{6}$
• D. $\cos \frac{\pi }{6}+i\sin \frac{\pi }{6}$

Answer 1

The correct option is D

$\cos \frac{\pi }{6}+i\sin \frac{\pi }{6}$

Explanation for the correct option.

$\begin{array}{rcl}x+\frac{1}{x}& =& \sqrt{3}\\ \Rightarrow x^2+1& =& \sqrt{3}x\\ & \Rightarrow & x^2-\sqrt{3}x+1=0\end{array}$

In the quadratic equation $\begin{array}{rcl}{x}^2-\sqrt{3}x+1& =& 0\end{array}$, $a=1,b=-\sqrt{3},\mathrm{and}c=1$

Roots of $\begin{array}{rcl}{x}^2-\sqrt{3}x+1& =& 0\end{array}$, will be

$$\begin{gathered} =\frac{\mathbf{-}\mathbf{b}\mathbf{\pm }\sqrt{{\mathbf{b}}^{\mathbf{2}}\mathbf{-}\mathbf{4}\mathbf{a}\mathbf{c}}}{\mathbf{2}\mathbf{a}} \\ =\frac{\sqrt{3}\pm \sqrt{3-4}}{2} \\ =\frac{\sqrt{3}\pm \sqrt{-1}}{2} \\ =\frac{\sqrt{3}\pm i}{2} \end{gathered}$$

$\frac{\sqrt{3}}{2}+\frac{1}{2}i=\cos \frac{\mathrm{\pi }}{6}+i\sin \frac{\mathrm{\pi }}{6}$

Hence, option D is correct.