Question

If $x + \dfrac{1}{x} = \sqrt 3$, then $x$ equals
1 $\cos \dfrac{\pi }{3} + i\sin \dfrac{\pi }{3}$
2 $\cos \dfrac{\pi }{2} + i\sin \dfrac{\pi }{2}$
3 $\sin \dfrac{\pi }{6} + i\cos \dfrac{\pi }{6}$
4 $\cos \dfrac{\pi }{6} + i\sin \dfrac{\pi }{6}$

Answer 1

To find the value of $x$ from the given trigonometric function, we need to apply trigonometric identity values with respect to its functions such that, by finding the final expression we can get the required value of $x$.
Formula used:
$\dfrac{ - b \pm \sqrt {{b^2} - 4ac}}{2a}$

Complete step by step answer:
Let us write the given data:
$x + \dfrac{1}{x} = \sqrt 3$
The given expression is written as:
$\Rightarrow {x^2} + 1 = \sqrt 3 x$
$\Rightarrow {x^2} - \sqrt 3 x + 1 = 0$ ………………………….. 1
As the obtained equation is of the form, $a{x^2} - bx + c$; hence, the roots from the obtained equation 1, by applying the formula we get:
$\dfrac{ - b \pm \sqrt {{b^2} - 4ac}}{2a}$
$\Rightarrow\dfrac{ \sqrt 3 \pm \sqrt {3 - 4}}{2}$
$\Rightarrow \dfrac{{\left( {\sqrt 3 \pm \sqrt {-1} } \right)}}{2}$
$\Rightarrow \dfrac{{\left( {\sqrt 3 \pm i} \right)}}{2}$
We know that, $\cos \dfrac{\pi }{6} = \dfrac{{\sqrt 3 }}{2}$ and $\sin \dfrac{\pi }{6} = \dfrac{1}{2}$, hence we get:
$\Rightarrow \dfrac{{\left( {\sqrt 3 \pm i} \right)}}{2} = \cos \dfrac{\pi }{6} + i\sin \dfrac{\pi }{6}$

So, the correct answer is “Option 4”.

Note: The key point to prove any trigonometric function is to note the chart of all related functions with respect to the equation, and must know all the trigonometric function values as with respect to its degrees, as we can find and solve the question easily.