Single Variable Calculus, Chapter 2, Review Exercises, Section Review Exercises, Problem 28

Use the Intermediate Value Theorem to show that $2 \sin x = 3 - 2x$ has a root in the interval $(0, 1)$

Let $f(x) = 2 \sin (x) - 3 + 2x$

Based from the definition of Intermediate Value Theorem, there exist a solution $c$ for the function between the interval $(a, b)$. Suppose that the function is continuous on that given interval. So, there exist a
number $c$ between and $1$ such that $f(x) = 0$ and that is, $f(c) = 0$




$\begin{equation} \begin{aligned} \qquad f(0) =& 2 \sin (0) - 3 + 2 (0) = -3\\ \\ \qquad f(1) =& 2 \sin (1) -3 + 2 (1) = 0.683 \end{aligned} \end{equation}$



By using Intermediate Value Theorem, we prove that

$\qquad$ if $0 < c < 1$ then $f(0) < f(c) < f(1)$

So,

$\qquad$ if $0 < c < 1$ then $-3 < 0 < 0.683$

Therefore,

$\qquad$ There exist a such solution $c$ for $2 \sin x = 3 - 2x$