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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 $x.$

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    $\begingroup$ Can you find $\cos x$ from the condition? $\endgroup$ – Daniel Fischer Mar 5 '16 at 11:07
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    $\begingroup$ $ -1<\cos x<1$. $\endgroup$ – Raio Mar 5 '16 at 11:09
  • $\begingroup$ What an answer ! $\endgroup$ – Jean Marie Mar 5 '16 at 11:11
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    $\begingroup$ To say $-1<\cos x<1$ you do not need to have such strong condition of that sum.. $\endgroup$ – user87543 Mar 5 '16 at 11:12
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    $\begingroup$ As @Daniel Fisher said, you have to concentrate on $\cos x$. Give it a name, $c$ for example. You are faced to a classical series $\sum_{n=0}^{\infty}c^n$. do you recognize it ? Do you now how to sum it up ? $\endgroup$ – Jean Marie Mar 5 '16 at 11:17
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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 $$ (u-1)(u^2-u-1)=0 $$

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HINT: consider the series $\sum_{n=-2}^\infty y^n$ with substitution $y=cosx$ and try to find $y$ s.t. $\sum_{n=-2}^\infty y^n = 8$.

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    $\begingroup$ How is this different from the hint that user Daniel Fischer has given? $\endgroup$ – user87543 Mar 5 '16 at 11:13
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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 $3\pi/5+2k\pi$, $-3\pi/5+2k\pi$

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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).

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Hint: If $-1 \lt \cos(x)\lt 1$ and $\cos(x)\ne 0$, we get that: $$ \begin{align} \sum_{n=-2}^{\infty} \cos(x)&=\frac{1}{\cos^2(x)}+\frac{1}{\cos(x)}+\frac{1}{1-\cos(x)}\\ &=\frac{1-\cos(x)+(1-\cos(x))\cos(x)+\cos^2(x)}{(1-\cos(x))\cos^2(x)}\\ &=\frac{1}{(1-\cos(x))\cos^2(x)} \end{align} $$

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    $\begingroup$ I think the term $1$ in the result of the series should not be there. $\endgroup$ – mickep Mar 5 '16 at 11:56

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