SOLUTION 7: We are given the equation $$x^3+2= \sin{x} \ \ \ \ \longrightarrow \ \ \ \ x^3+2-\sin{x}=0$$ Let function $$f(x)=x^3+2-\sin{x} \ \ \ \ and \ choose \ \ \ \ m=0$$ This function is continuous for all values of $x$ since it is the DIFFERENCE of continuous functions; it is well-known that $\sin{x}$ is continuous for all values of $x$ and $x^3+2$ is continuous for all values of $x$ since it is a polynomial. To establish an appropriate interval consider the graph of this function. (Please note that the graph of the function is not necessary for a valid proof, but the graph will help us understand how to use the Intermediate Value Theorem. On many subsequent problems, we will solve the problem without using the "luxury" of a graph.)

Note that $$f(0)= (0)^3+2-\sin{(0)} =2>0 \ \ \ \ and \ \ \ \ f(-\pi)= (-\pi)^3+2-\sin{(-\pi)}= -\pi^3+2-0=2-\pi^3 \approx -29<0$$
so that $$f(-\pi) \approx -29 < m <2=f(0)$$
i.e., $m=0$ is between $f(-\pi)$ and $f(0)$. Now choose the interval to be $\ [-\pi, 0]$.

The assumptions of the Intermediate Value Theorem have now been met, so we can conclude that there is some number $c$ in the interval $[-\pi, 0]$ which satisfies $$f(c)=m$$ i.e., $$c^3+2-\sin{c}=0$$ and the equation is solvable.
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}