Answer
The equation has 3 solutions: $(-1.895, 0, 1.895)$.
Work Step by Step
$$2\sin x=x$$ $$2\sin x-x=0$$
To prove the equation has a solution using the Immediate Value Theorem, we need to find 2 values $x_1$ and $x_2$ and show that there is a change of sign in the function $f(x)=2\sin x-x$ as $x$ goes from $x_1$ to $x_2$.
- Take $x_1=\pi$: $f(\pi)=2\sin\pi-\pi=2\times0-\pi=-\pi\lt0$
- Take $x_2=-\pi$: $f(-\pi)=2\sin(-\pi)-(-\pi)=2\times0+\pi=\pi\gt0$
So there is a change of sign in $f(x)$ as $x$ goes from $-\pi$ to $\pi$.
- On $[-\pi,\pi]$: $\lim_{x\to c}f(x)=\lim_{x\to c}(2\sin x-x)=2\sin c-c=f(c)$
So $f(x)$ is continuous on $[-\pi,\pi]$.
Therefore, according to the Intermediate Value Theorem, there must be a value $x=c\in[-\pi,\pi]$ such that $f(c)=0$. In other words, the equation $2\sin x-x=0$ or $2\sin x=x$ has at least one solution in the interval $[-\pi,\pi]$.
The graph of the function $f(x)=2\sin x-x$ is enclosed below. Looking at the graph, it turns out that the curve $f(x)$ crosses the line $y=0$ at $3$ points. The $x$-coordinates of these points are the solutions of the equation.
So the equation has 3 solutions: $(-1.895, 0, 1.895)$
