Answer
Turn the equation into the function $f(x)=\cos x-x$ and apply the Intermediate Value Theorem for Root Finding.
Work Step by Step
$$\cos x=x$$ $$\cos x-x=0$$
To prove the equation has at least one solution, we can apply the Intermediate Value Theorem for Root Finding, which requires to us to do the followings:
- Take the function $f(x)=\cos x-x$
- Find 2 values $x_1$ and $x_2$ such that $f(x_1)\lt0$ while $f(x_2)\gt0$ or vice versa.
- Show that $f(x)$ is continuous on $[x_1, x_2]$.
Considering the function $f(x)=\cos x-x$:
- Domain: $(-\infty,\infty)$
1) Continuity:
On $(-\infty,\infty)$, we notice that $\lim_{x\to c}(\cos x-x)=\cos c-c=f(c)$
That means $f$ is continuous on $(-\infty,\infty)$.
2) Find $x_1$ and $x_2$:
Here I graph $f(x)=\cos x-x$ for better findings, which I have also enclosed below.
- At $x_1=0$: $f(0)=\cos0-0=1-0=1\gt0$
- At $x_2=\pi/2$: $f(\pi/2)=\cos(\pi/2)-(\pi/2)=0-\pi/2=-\pi/2\lt0$
And since $f$ is continuous on $R$, it is continuous on $[0,\pi/2]$
Therefore, according to the Intermediate Value Theorem, there must exist at least a value of $x=c\in[0,\pi/2]$ such that $f(c)=0$.
In other words, the equation $\cos x-x=0$ or $\cos x=x$ has at least one solution in $[0,\pi/2]$.
