<s>[INST] Roots of Taylor's series. 
Show that there is exactly one value of x which satisfies the equation $$2\cos^2 (x^3+x)=2^x+2^{-x} $$

I solved this using Taylor's series:
$$2^x+2^{-x}=2\{1+\frac {x^2 \{\ln2\}^2}{2!}+\frac {x^4 \{\ln2\}^4}{4!}....\} $$
$$\cos^2 (x^3+x)=1+\cos (2x^3+2x)=2-\frac{{2x^3+2x}^2}{2!}.... $$
Equating the two gives 
$$ x^2\{{\ln2}^2+.....\}$$
$$\implies x=0 $$ 
But I can't seem to be able to find an argument justifying the existence if only one root. If I said that as the other term is an infinite series hence it's roots are undefined,  would that be correct? 
 [/INST] As for real $x, 2^{\pm x}>0$
$$\frac{2^x+2^{-x}}2\ge \sqrt{2^x\cdot 2^{-x}}=1\  \ \ \ (1)$$
So, we have $$2\cos^2(x^3+x)=2^x+2^{-x}\ge2$$
$$\iff \sin^2(x^3+x)\le0$$
For real $x,\displaystyle\sin^2(x^3+x)\ge0\implies$ the relation will hold iff 
$\displaystyle2^x+2^{-x}=2$ and $\displaystyle\sin^2(x^3+x)=0\iff\sin(x^3+x)=0$
But, the equality in $(1)$ will hold iff $\displaystyle2^x=2^{-x}\iff 2^{2x}=1$
Can you take it home from here?

Another way:
If $\displaystyle a+a^{-1}=2\cos\theta, a^2-2a\cos\theta+1=0$
$\displaystyle\implies a=\cos\theta\pm i\sin\theta$
If $a$ is real, $\sin\theta$ must be $\displaystyle0\implies a=\cos\theta=\pm1$
Again, if $\displaystyle a>0, a=1\implies \theta=2n\pi$ where $n$ is any integer
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