How find this $\sum_{i=0}^{5}\frac{1}{2+\cos{\left(x+\frac{i\pi}{3}\right)}}\cdot \frac{1}{2+\cos{\left(x+\frac{(i+1)\pi}{3}\right)}}$ Find this follow  function $f(x)$ range ,where $x\in R$,
$$f(x)=\sum_{i=0}^{5}\dfrac{1}{2+\cos{\left(x+\dfrac{i\pi}{3}\right)}}\cdot \dfrac{1}{2+\cos{\left(x+\dfrac{(i+1)\pi}{3}\right)}}$$
or
$$f(x)=\dfrac{1}{2+\cos{x}}\cdot\dfrac{1}{2+\cos{\left(x+\dfrac{\pi}{3}\right)}}+\dfrac{1}{2+\cos{\left(x+\dfrac{\pi}{3}\right)}}\cdot\dfrac{1}{2+\cos{\left(x+\dfrac{2\pi}{3}\right)}}+\dfrac{1}{2+\cos{\left(x+\dfrac{2\pi}{3}\right)}}\cdot\dfrac{1}{2+\cos{\left(x+\dfrac{3\pi}{3}\right)}}+\dfrac{1}{2+\cos{\left(x+\dfrac{3\pi}{3}\right)}}\cdot\dfrac{1}{2+\cos{\left(x+\dfrac{4\pi}{3}\right)}}+\dfrac{1}{2+\cos{\left(x+\dfrac{4\pi}{3}\right)}}\cdot\dfrac{1}{2+\cos{\left(x+\dfrac{5\pi}{3}\right)}}+\dfrac{1}{2+\cos{\left(x+\dfrac{5\pi}{3}\right)}}\cdot\dfrac{1}{2+\cos{\left(x+\dfrac{6\pi}{3}\right)}}$$
I think $f(x)$ have simple form,becasue this is exam problem
I think maybe can  use
$$f(2\pi-x)+f(x)=?$$
or maybe have use this
$$\cos{x}\cos{y}=\dfrac{1}{2}[\cos{(x-y)}+\cos{(x+y)}]$$
I use  this two idea all solve this problem,
Thank you 
 A: HINT
I should start simplifying the sum of the first and the sixth terms (result = A), then the sum of the second and fifth terms (result = B), then the sum of the third and fourth terms (result = C). Now, I should simplify A + B (result = D) and finally simplify C + D.    
You will arrive to a surprizingly simple result.
A: $$\text{Let }y_r=\left(2+\cos\left(x+\frac{r\pi}3\right)\right)\left(2+\cos\left(x+\dfrac{(r+1)\pi}3\right)\right)$$
$$y_r=4+2\left[\cos\left(x+\frac{r\pi}3\right)+\cos\left(x+\dfrac{(r+1)\pi}3\right)\right]+\cos\left(x+\frac{r\pi}3\right)\cos\left(x+\dfrac{(r+1)\pi}3\right)$$
Applying $2\cos A\cos B$ and $\cos C+\cos D$      formula,
$$2y_r=8+8\cos\left(x+\frac{(2r+1)\pi}6\right)\cos\frac\pi6+\left[\cos\frac\pi3+\cos\left(2x+\frac{(2r+1)\pi}3\right)\right]$$
$$=8+4\sqrt3c+\frac12+2c^2-1\text{ where }c=\cos\left(x+\frac{(2r+1)\pi}6\right)$$
$$\implies 4y_r=4c^2+8\sqrt3c+15\ \ \ \  (1)$$
So, we need to find $\displaystyle\sum_{r=0}^5\frac1{y_r}=\frac{\sum_{\text{cyc}} 5 y_r\text{-s at a time}}{\prod _{r=0}^5 y_r}$ which can be easily managed by Vieta's formula
Now,  $\displaystyle \cos6\left(x+\frac{(2r+1)\pi}6\right)=\cos\left(6x+\overline{2r+1}\pi\right)=-\cos6x$
So if $\displaystyle \cos6A=-\cos6x=\cos(\pi+6x),$
$\displaystyle\implies6A=2n\pi\pm(\pi+6x)$ where $n$ is any integer
$\displaystyle\implies A=x+\frac{(2n+1)\pi}6$ where $0\le n\le 5\  \  \ \ (2)$
So, as I've explained in the comment, 
$(2)$ is the set of roots of $\displaystyle32c^6−48c^4+18c^2+\cos6A-1=0\ \ \ \ (3)$
Now from $\displaystyle (1),c^2=\frac{4y_r-8\sqrt3c-15}{15}\  \ \  \ (4)$
Squaring we get $\displaystyle c^4=\frac{16y_r^2+192c^2+225-(64\sqrt3y_r)c-120y_r+240\sqrt3c}{225} \ \ \ \ (5)$
Put the value of $c^2$  from $(4)$ in $(5)$
Now multiply $(4),(5)$ to get the value of $c^6$
Put the values of $c^6,c^4,c^2$ in $(3)$ to find $c$ in terms of $y_r$
Now put the values of $c,c^2$ in $(4)$ to form a Sextic equation in $y_r$ where we need to apply Vieta's formula mentioned above
