Why we use $2\cos\theta=\sqrt3$ to find $x^{72}+x^{66}$ given that $x+\frac1x=\sqrt3$? $x + \frac{1}{x} = \sqrt 3$  then find the value of $x^{72}+x^{66}$.
My teacher said that in this case, where $x + \frac{1}{x}$ is equal to any value which is less than $2$, like $1$ or $\sqrt{3}$, just put this value in equal of $2 \cos \theta$ and after you will get value of $\theta$ after that divide the value $180$ by that theta value and after getting the result put that result  in equation $x^n + 1=0$,  as "$n$". 
Please explain what is basic concept behind this.

If 
$$x+\dfrac{1}{x} = 2\cos\theta$$ 
then 
$$x^{n}+\dfrac{1}{x^n} = 2\cos(n\theta)$$
A way to prove this is with a concept known as mathematical induction, which you may not be ready to do/understand yet. If you wanted to read up on it, attempt it, it would be a little easier to prove the equivalent: If $x^2+1 = 2x\cos\theta$ then $x^N+1 = x^{2n}+1 = 2x^n\cos(n\theta)$.
$$x^{72}+x^{66} = x^{66}(x^6+1)$$
So we are interested in the case where $N=6$.
If $x+\dfrac{1}{x}=\sqrt{3}$ then we solve 
$$2\cos(\theta) = \sqrt{3} \iff \cos(\theta) = \dfrac{\sqrt{3}}{2}$$
So $\theta = 30^o$ (there are other possibilities). I suspect the smallest acute angle is what is desired (I really don't understand the necessity of the whole $180^o$ thing, I thought my previous answer was neater and succinct). 
$\dfrac{180^o}{30^o} = 6 = N = 2n$.
Since $x^{2}+1 = 2\cos\theta$ implies $x^{2n}+1 = 2x^n\cos(n\theta)$ we have
$$x^6 + 1 = 2x^3\cos(3\cdot 30^o) = 2x^3\cdot\cos(90^o) = 2x^3(0)=0$$
$$x^{72}+x^{66} = x^{66}(x^6+1) = x^{66}(0)=0$$
