$x + \frac{1}{x} = \sqrt 3$ 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.
On
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
$$x^6 + 1 = 2x^3\cos(3\cdot 30^o) = 2x^3\cdot\cos(90^o) = 2x^3(0)=0$$
$$x^{72}+x^{66} = x^{66}(x^6+1) = x^{66}(0)=0$$
On
Possibly what your teacher means is this:
Let $x=e^{i\theta}$ so that$$x+\frac 1x=e^{i\theta}+e^{-i\theta}=2\cos\theta=\sqrt{3}$$
Therefore $$\theta=\pm\frac{\pi}{6}+n.2\pi$$
Note that $30=\frac{180}{\color{red}{6}}$
So $$x^{72}+x^{66}=x^{66}(x^{\color{red}{6}}+1)=x^{66}(-1+1)=0$$
On
Consider $$ x+\frac{1}{x}=a $$ with $-2< a< 2$, and set $a=2\cos\theta$. There are two values of $\theta$ in $[0,2\pi)$ satisfying this relation, call them $\theta_1$ and $\theta_2$. Then $x_1=\cos\theta_1+i\sin\theta_1$ and $x_2=\cos\theta_2+i\sin\theta_2$ are distinct roots of the equation, which is of degree two, so they are the roots.
For $a=\pm2$ the discriminant is zero, so the equation has a double root, with just one value for $\theta$.
In your case, $\theta=\pm\frac{\pi}{6}$, so $x^6=-1$.
On
Maybe the teacher meant that you just follow his instructions and don't ask questions.
Otherwise, the "basic concept behind this" can only be explained in terms of complex numbers. As others have remarked an $x$ satisfying the given equation has to be complex. We therefore may write $$x=r(\cos\theta+i\sin\theta)$$ with $r\geq0$ and $\sin\theta\ne0$. Then ${1\over x}={1\over r}(\cos\theta-i\sin\theta)$, so that we have to satisfy $$\sqrt{3}=x+{1\over x}=\left(r+{1\over r}\right)\cos\theta+i\left(r-{1\over r}\right)\sin\theta\ .$$ As $\sqrt{3}$ is real it follows that $r-{1\over r}=0$, hence $r=1$, and then that $2\cos\theta=\sqrt{3}$, or $$\theta=\pm30^\circ=\pm{\pi\over6}\quad({\rm modulo}\ 2\pi)\ .$$ From this we deduce that $$x^6=\cos(6\theta)+i\sin(6\theta)=\cos\pi+i\sin\pi=-1\ ,$$ and this implies $$x^{72}+x^{66}=(-1)^{12}+(-1)^{11}=1+(-1)=0\ .$$
We have to solve the equation $\ x+\frac{1}{x}=a$. This leads to the equation
$$x^2-ax+1=0$$
If $\ 0\le a<2$, the solutions must be complex.
If $u+vi$ is one of the solutions, the other one is $u-vi$. So, the absolute values of the solutions are equal and multiply to $1$. Therefore, every solution of this equation has absolute value $1$.
If the angle that is formed by a solution is a divisor of $180$ , we have $k\cdot\theta=180$. That means, every solution of the equation satisfies $x^k=-1$.
In the given example, the angle is $30°$, that means $k=6$, so every solution satisfies $x^6=-1$
This immediately shows $x^{72}+x^{66}=0$
What remains is : Why do we take $\ 2cos\theta=a$ ?
The sum of the solutions must be $a$, so the real part of both solutions is $u=\frac{a}{2}$. The absolute value of the solutions is $1$, so we have $u=cos(\theta)$ and therefore $a=2cos(\theta)$.