# 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 $x.$

• Can you find $\cos x$ from the condition? – Daniel Fischer Mar 5 '16 at 11:07
• $-1<\cos x<1$. – Raio Mar 5 '16 at 11:09
• What an answer ! – Jean Marie Mar 5 '16 at 11:11
• To say $-1<\cos x<1$ you do not need to have such strong condition of that sum.. – user87543 Mar 5 '16 at 11:12
• 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 ? – Jean Marie Mar 5 '16 at 11:17

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 $$(u-1)(u^2-u-1)=0$$

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

• How is this different from the hint that user Daniel Fischer has given? – user87543 Mar 5 '16 at 11:13

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 $3\pi/5+2k\pi$, $-3\pi/5+2k\pi$

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).

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}

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