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**Hint:**In the given question, we can see that we are given a series of sine functions having angles forming an arithmetic progression. Now, we are given the value of the sum of the terms of the series and we are asked to find the total number of terms of the series.

**Complete step-by-step solution:**

According to the given question, we can clearly see that we are given a series of n-1 terms and the angles which are taken in sine series are forming an arithmetic progression which is $S=\sin \dfrac{\pi }{n}+\sin \dfrac{2\pi }{n}+\sin \dfrac{3\pi }{n}+\cdots +\sin \dfrac{\left( n-1 \right)\pi }{n}+\sin \dfrac{n\pi }{n}$.

Now, we know that the formula to find the sum of the series when the angles are given in arithmetic progression is $\sum\limits_{k=0}^{k=n-1}{\sin \left( a+kd \right)}=\dfrac{\sin \left( \dfrac{nd}{2} \right)}{\sin \left( \dfrac{d}{2} \right)}\sin \left( a+\left( n-1 \right)\dfrac{d}{2} \right)$ .

Now, in order to use this formula, the values of different variables given in the question are as follows:

$a=\dfrac{\pi }{n}$ , $d=\dfrac{\pi }{n}$ , $n=n-1$ where n denotes the number of terms of the series and a is representing the first term of the progression and d is the common difference of the progression.

We have taken n-1 terms as last term is 0 as last term is $\sin \pi $after cancelling the n from the numerator and denominator.

Now, putting all these value in the formula we get,

$\dfrac{\sin \left[ \left( n-1 \right)\dfrac{\pi }{2n} \right]}{\sin \left( \dfrac{\pi }{2n} \right)}\sin \left(\dfrac{\pi }{n}+\left( n-2 \right)\dfrac{\pi }{2n}\right)$

Now, simplifying this we get,

$\begin{align}

& \dfrac{\sin \left[ \left( n-1 \right)\dfrac{\pi }{2n} \right]\sin \left( \dfrac{2\pi +n\pi -2\pi }{2n} \right)}{\sin \dfrac{\pi }{2n}} \\

&\Rightarrow\dfrac{\sin \left[ \left( n-1 \right)\dfrac{\pi }{2n} \right]}{\sin \left( \dfrac{\pi }{2n} \right)} \\

\end{align}$

As we know that $\sin \dfrac{\pi }{2}=1$ .

Further, we get $\dfrac{\sin \left( \dfrac{n\pi }{2n}-\dfrac{\pi }{2n} \right)}{\sin \left( \dfrac{\pi }{2n} \right)}$ and now we know that the sum is equal to $2+\sqrt{3}$ .

Now,

$\begin{align}

&\Rightarrow \dfrac{\sin \left( \dfrac{n\pi }{2n}-\dfrac{\pi }{2n} \right)}{\sin \left( \dfrac{\pi }{2n} \right)} \\

& \Rightarrow \dfrac{\sin \left( \dfrac{\pi }{2}-\dfrac{\pi }{2n} \right)}{\sin \left( \dfrac{\pi }{2n} \right)}=2+\sqrt{3} \\

\end{align}$

And we know that $\sin \left( \dfrac{\pi }{2}-\theta \right)=\cos \theta $ , therefore, using this we get

$\begin{align}

& \dfrac{\cos \left( \dfrac{\pi }{2n} \right)}{\sin \left( \dfrac{\pi }{2n} \right)}=2+\sqrt{3} \\

& \Rightarrow \cot \left( \dfrac{\pi }{2n} \right)=2+\sqrt{3} \\

\end{align}$ .

Now, we know that $\cot {{15}^{\circ }}=\dfrac{\sqrt{3}+1}{\sqrt{3}-1}$ and rationalising this we get $\dfrac{\sqrt{3}+1}{\sqrt{3}-1}\times \dfrac{\sqrt{3}+1}{\sqrt{3}+1}=2+\sqrt{3}$

And this implies that

$\begin{align}

& 15\times \dfrac{\pi }{180}=\dfrac{\pi }{2n} \\

& \Rightarrow \dfrac{1}{12}=\dfrac{1}{2n} \\

& \Rightarrow n=6 \\

\end{align}$

**Therefore, the number of terms of the series is 6.**

**Note:**In such a type of question, we need to observe the given series very carefully as it involves many types of forms in itself, we just need to identify the requirement of the question and proceed similarly. Also, we need to be aware of the concept of the arithmetic progression.

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