How do I prove or disprove that the equation $\cos({\sin{x}})=1/x$ has only one real root? 
How do I prove or disprove that the equation $\cos({\sin{x}})=1/x$ has only one real root?

It's easy to do graphically, but I'm more interested in the algebraic method.
My attempt:
First, we can show that $\cos (1)=0.5403023 \ldots \leq \cos (\sin x) \leq 1$. This implies that the root(s) must be such that $x \in[1 ; 1 / \cos (1)=1.8508157 \ldots]$.
Second, we can show that if $x \in[1 ; \pi / 2]$ we have $\cos (\sin x)<1 / x$. This implies that the root(s) must be such that $x \in[\pi / 2 ; 1 / \cos (1)]$.
Finally, we show that in this interval the function $\cos({\sin{x}})$ is increasing but the function $1/x$ is decreasing, and that implies that there is only one root.
Finding the root of $f(x)=\cos (\sin x)-1 / x$ can be done using the derivative $f^{\prime}(x)=-(\cos x) \sin (\sin x)+1 / x^{2}$.
 A: It is easy to show that $\cos\sin x=1/x$ has no solution over $(0,1)$ as $|\cos t|\le1$ and $1/t>1$. Thus we can transform the equation to $f(x)=\sin x-\arccos(1/x)=0$ which has domain $[1,\infty)$. Since $\arccos(1/x)$ has derivative $1/(x\sqrt{x^2-1})$, it is a strictly increasing function with range $[0,\pi/2)$. Furthermore, as $\arccos(1/2)=\pi/3>1$, we know that any roots must lie in $[1,2]$, and in particular, exactly one of them is in $[\pi/2,2]$ as $\sin x$ is decreasing.
It remains to show that $f(x)>0$ in $[1,\pi/2)$. As $f(x)$ is positive at the endpoints it suffices to show that $f'(x)=\cos x-1/(x\sqrt{x^2-1})<0$. Since $1/(x\sqrt{x^2-1})$ has derivative $-(2x^2-1)/(x^2(x^2-1)^{3/2})$, it is a strictly decreasing function, so we have the upper bound $f'(x)<\cos x-1/((\pi/2)\sqrt{\pi^2/4-1})$. This is negative when $x\in(a,\pi/2)$ where $a=\arccos(4/(\pi\sqrt{\pi^2-4}))$.
Finally to show that $f'(x)<0$ in $[1,a]$ (and hence $f(x)>0$), we use the same trick on the shorter endpoint. In this interval, we have the upper bound $f'(x)<\cos x-1/(a\sqrt{a^2-1})$ which is indeed negative. Therefore $f(x)$ has no roots in $[1,\pi/2)$, and the conclusion follows.
A: For nonzero $x\in\mathbb{R}$, let
$$
\left\lbrace
\begin{align*}
f(x)&=\cos(\sin(x))\\[4pt]
g(x)&=\frac{1}{x}\\[4pt]
h(x)&=f(x)-g(x)\\[4pt]
\end{align*}
\right.
$$
and let
$$
A={\bigl{\{}}a\in\mathbb{R}{\,{\large{\mid}}\,}h(a)=0{\bigr{\}}}
$$
Claim:$\;|A|=1$.

Proof:

On the interval $\bigl[{\large{\frac{\pi}{2}}},\pi\bigr)$,

*

*$f$ is strictly increasing.$\\[4pt]$

*$g$ is strictly decreasing.

hence, since
$$
h\Bigl({\small{\frac{\pi}{2}}}\Bigr) < 0 < h(\pi)
$$
it follows that
$$
\Bigl|A\cap \bigl[{\small{\frac{\pi}{2}}},\pi\bigr)\Bigr|=1
$$
Now let $a\in A$.

To show $|A|=1$, it suffices to show $a\in\bigl[{\large{\frac{\pi}{2}}},\pi\bigr)$.
\begin{align*}
\text{Then}\;\;&
-1\le \sin(a)\le 1
\\[4pt]
\implies\;&
0 < \cos(\sin(a))\le 1
\\[4pt]
\implies\;&
0 < \frac{1}{a}\le 1
\\[4pt]
\implies\;&
a\ge 1
\\[4pt]
\end{align*}
Next, suppose $a\in\bigl[1,{\large{\frac{4}{3}}}\bigr)$.

On the interval $\bigl[1,{\large{\frac{4}{3}}}\bigr)$,

*

*$f$ is strictly decreasing.$\\[4pt]$

*$g$ is strictly decreasing.

so then
\begin{align*}
&
\cos(\sin(1)) < {\small{\frac{3}{4}}}
\\[4pt]
\implies\;&
f(1) < g\Bigl({\small{\frac{4}{3}}}\Bigr)
\\[4pt]
\implies\;&
f(a) < g(a)
\\[4pt]
\end{align*}
contradiction, hence $a\ge{\large{\frac{4}{3}}}$.

Next, suppose $a\in\bigl[{\large{\frac{4}{3}}},{\large{\frac{\pi}{2}}}\bigr)$.

On the interval $\bigl[{\large{\frac{4}{3}}},{\large{\frac{\pi}{2}}}\bigr)$,

*

*$f$ is strictly decreasing.$\\[4pt]$

*$g$ is strictly decreasing.

so then
\begin{align*}
&
\cos\Bigl(\sin\Bigl({\small{\frac{4}{3}}}\Bigr)\Bigr) < {\small{\frac{2}{\pi}}}
\\[4pt]
\implies\;&
f\Bigl({\small{\frac{4}{3}}}\Bigr) < g\Bigl({\small{\frac{\pi}{2}}}\Bigr)
\\[4pt]
\implies\;&
f(a) < g(a)
\\[4pt]
\end{align*}
contradiction, hence $a\ge{\large{\frac{\pi}{2}}}$.

Finally, we have
\begin{align*}
&
\cos(\sin(a))\ge\cos(1)
\\[4pt]
\implies\;&
f(a) \ge \cos(1)
\\[4pt]
\implies\;&
f(a) > {\small{\frac{1}{\pi}}}
\\[4pt]
\implies\;&
g(a) > {\small{\frac{1}{\pi}}}
\\[4pt]
\implies\;&
{\small{\frac{1}{a}}} > {\small{\frac{1}{\pi}}}
\\[4pt]
\implies\;&
a < \pi
\\[4pt]
\end{align*}
hence we have $a\in\bigl[{\large{\frac{\pi}{2}}},\pi\bigr)$, which completes the proof.
