If $x = a(2\theta - \sin 2\theta )$ and $y = a(1 - \cos 2\theta )$, find $\dfrac{{dy}}{{dx}}$ when $\theta = \dfrac{\pi }{3}$?

Hint: In this question, we are given two equations- one in the terms of $x$ and $\theta $, other in the terms of $y$ and $\theta $ , and we have been asked to find $\dfrac{{dy}}{{dx}}$. For this, we either need the equations in the terms of $y$ and $x$, or we can first differentiate the 2 equations separately and then divide them in the form- \[\dfrac{{\dfrac{{dy}}{{d\theta }}}}{{\dfrac{{dx}}{{d\theta }}}}\] to find $\dfrac{{dy}}{{dx}}$ . After this, we are given a value of $\theta $. We have to put this value in $\dfrac{{dy}}{{dx}}$ and find the required answer.

Formula used: 1) $\cos 2\theta = 1 - 2{\sin ^2}\theta $
2) $\sin 2\theta = 2\sin \theta \cos \theta $
3) $\cot \dfrac{\pi }{3} = \dfrac{1}{{\sqrt 3 }}$

Complete step-by-step answer:
We are given two different equations. First, we will differentiate both of them individually. And then we will find $\dfrac{{dy}}{{dx}}$ at $\theta = \dfrac{\pi }{3}$.
Equation 1:
$ \Rightarrow x = a(2\theta - \sin 2\theta )$
Differentiating with respect to $\theta $,
$ \Rightarrow \dfrac{{dx}}{{d\theta }} = a(2 - 2\cos 2\theta )$
Taking 2 common,
$ \Rightarrow \dfrac{{dx}}{{d\theta }} = 2a(1 - \cos 2\theta )$
Using trigonometric identity- $\cos 2\theta = 1 - 2{\sin ^2}\theta $
$ \Rightarrow \dfrac{{dx}}{{d\theta }} = 2a(2{\sin ^2}\theta )$
Simplifying,
$ \Rightarrow \dfrac{{dx}}{{d\theta }} = 4a{\sin ^2}\theta $
Equation 2:
$y = a(1 - \cos 2\theta )$
Differentiating with respect to $\theta $,
$ \Rightarrow \dfrac{{dy}}{{d\theta }} = a(2\sin 2\theta )$
$ \Rightarrow \dfrac{{dy}}{{d\theta }} = 2a\sin 2\theta $
Using formula- $\sin 2\theta = 2\sin \theta \cos \theta $
$ \Rightarrow \dfrac{{dy}}{{d\theta }} = 4a\sin \theta \cos \theta $
To find $\dfrac{{dy}}{{dx}}$, divide $\dfrac{{dy}}{{d\theta }}$ and $\dfrac{{dx}}{{d\theta }}$.
$ \Rightarrow \dfrac{{\dfrac{{dy}}{{d\theta }}}}{{\dfrac{{dx}}{{d\theta }}}} = \dfrac{{4a\sin \theta \cos \theta }}{{4a{{\sin }^2}\theta }}$
Simplifying,
$ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{\cos \theta }}{{\sin \theta }}$
We know, $\dfrac{{\cos \theta }}{{\sin \theta }} = \cot \theta $.
Therefore, $\dfrac{{dy}}{{dx}} = \cot \theta $
Now, we put $\theta = \dfrac{\pi }{3}$
$ \Rightarrow \dfrac{{dy}}{{dx}} = \cot \dfrac{\pi }{3} = \dfrac{1}{{\sqrt 3 }}$

Therefore $\dfrac{1}{{\sqrt 3 }}$ is the required answer.

Note: To find $\dfrac{{dy}}{{dx}}$ of the equations which are not in the terms of y and x, we have to first differentiate the equations in whichever form they are and then divide them in such a way that they give us $\dfrac{{dy}}{{dx}}$.
For example: If an equation is in the terms of y and t, first we find $\dfrac{{dy}}{{dt}}$. Then, if another equation is in the form of x and t, we find $\dfrac{{dx}}{{dt}}$. After we have differentiated both the questions, we divide them in the way- \[\dfrac{{\dfrac{{dy}}{{d\theta }}}}{{\dfrac{{dx}}{{d\theta }}}}\]. This will give us $\dfrac{{dy}}{{dx}}$.