# From proof to simpler proof

Solve $\sin(x)=2x$ for $x$.

By mere inspection, we have $x=0$.

Visually, it appears that $0$ is the only solution (see Fig.1 or Fig. 2)

Fig. 1

Fig. 2

To show that $0$ is the only solution of $\sin(x)=2x$ analytically, let $f(x) = \sin(x)-2x$ $0$ is a solution of $\sin(x)=2x$ means $f(0) = 0\quad\quad\quad(1)$

Suppose there is a solution $x^* \neq 0\quad\quad\quad(2)$

then, $f(x^*)=\sin(x^*)-2x^*=0\quad\quad\quad(3)$

Since $f(x)$ is an function continuous and differentiable on $(-\infty, +\infty)$,

by Lagrange’s Mean-Value Theorem (see “A Sprint to FTC“), there $\exists c \in (0, x^*)$ such that $f(x^*)-f(0)=f'(c)(x^*-0)$ $\overset{(1), (3)}{\implies} 0 = f'(c) x^*\quad\quad\quad(4)$ $\overset{(2)}{\implies} f'(c)=0\quad\quad\quad(5)$.

We know $f'(x) = \cos(x)-2$.

From (5), we have $\cos(c)-2=0$.

i.e., $\cos(c)=2$.

This is not possible since $\forall x \in R, -1 \le \cos(x) \le 1$.

A simpler alternative without direct applying Lagrange’s Mean-Value Theorem is: $f(x) = \sin(x)-2x \implies f'(x) = \cos(x) - 2 \overset{-1 \le \cos(x) \le 1}{\implies} f'(x) < 0\implies$ $f(x) =\sin(x)-2x$ is a strictly decreasing function.

Since $f(0) = 0, \forall x<0, f(x) > 0$ and $\forall x>0, f(x) < 0$.

Therefore, $0$ is the only solution of $f(x)=0$. i.e., $0$ is the only solution of $\sin(x) = 2x$.