If $\sum_{n=-2}^{\infty}\cos^n x=8$, then find $x.$ Let $0<2x<\pi$. If $$\sum_{n=-2}^{\infty}\cos^n x=8,$$ then please find $x.$ 
I tried $\sum_{n=-2}^{\infty}\cos^n x=\frac{1}{1-\cos x}+\frac{1}{\cos x}+\frac{1}{\cos^2 x}=\frac{1+\cos x}{\cos^2 x \sin^2 x}=8.$ But I cant find $x.$
 A: Since $0<x<\pi/2$, you know that $0<\cos x<1$, so the series
$$
\sum_{n=-2}^{\infty}(\cos x)^n=\sum_{m=0}^{\infty}(\cos x)^{m-2}=
\frac{1}{\cos^2x}\sum_{m=0}^{\infty}(\cos x)^{m}
$$
converges to
$$
\frac{1}{\cos^2x}\frac{1}{1-\cos x}
$$
Solving
$$
\frac{1}{t^2(1-t)}=8
$$
shouldn't be difficult as it transforms into
$$
t^3-t^2+\frac{1}{8}=0
$$
and it is perhaps easier setting $t=u/2$, so the equation can be rewritten as
$$
u^3-2u^2+1=0
$$
where you can spot the root $u=1$; after division you find
$$
(u-1)(u^2-u-1)=0
$$
A: HINT: consider the series $\sum_{n=-2}^\infty y^n$ with substitution $y=cosx$ and try to find $y$ s.t. $\sum_{n=-2}^\infty y^n = 8$.
A: Hint: If $-1 \lt \cos(x)\lt 1$ and $\cos(x)\ne 0$, we get that:
$$
\begin{align}
\sum_{n=-2}^{\infty} \cos(x)&=\frac{1}{\cos^2(x)}+\frac{1}{\cos(x)}+\frac{1}{1-\cos(x)}\\
&=\frac{1-\cos(x)+(1-\cos(x))\cos(x)+\cos^2(x)}{(1-\cos(x))\cos^2(x)}\\
&=\frac{1}{(1-\cos(x))\cos^2(x)}
\end{align}
$$
A: Substituting y to cos(x), we have $\sum_{n=-2}^\infty y^n=-\frac{1}{y^2(y-1)}=8.$
The solutions are $1/2$, $\frac{1-\sqrt{5}}{4}$, $\frac{1+\sqrt{5}}{4}$
Then we have $\cos(x)=1/2$ so $x = \pi/3+2k\pi$ or $x=-\pi/3+2k\pi$ ($k \in \mathbb{Z}$)
the two other solutions give four values for x.
$\pi/5+2k\pi$, $-\pi/5+2k\pi$ and $3\pi/5+2k\pi$, $-3\pi/5+2k\pi$
A: You can continue your calculation, to write
$$
\frac{1+\cos x}{\cos^2x\sin^2x}=\frac{1+\cos x}{\cos^2x(1-\cos^2x)}=\frac{1}{\cos^2x(1-\cos x)}.
$$
Next, let $y=\cos x$. Then you should solve the equation
$$
\frac{1}{y^2(1-y)}=8,\quad\text{i.e.}\quad y^3-y^2+\frac{1}{8}=0.
$$
To solve this equation in $y$, I think you can guess one (rational) root. The other ones will come from a quadratic equation after polynomial division. Finally, solve the equation $\cos x=y$ for the different $y$ you got. Don't forget to only include the $x$'s in the interval $0<x<\pi/2$ (I get two such).
